Circles

Let us consider a circle with center O and radius 8 cm.

The diagram is given as:

Consider a point A 17 cm away from the center such that OA = 17 cm

A tangent is drawn at point A on the circle from point B such that OB = radius = 8 cm

To Find: Length of tangent AB = ?

As seen OB ⏊ AB

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

∴ In right - angled ΔAOB, By Pythagoras Theorem

[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]

(OA)² = (OB)² + (AB)²

(17)² = (8)² + (AB)²

289 = 64 + (AB)²

(AB)² = 225

AB = 15 cm

∴ The length of the tangent is 15 cm.

Let us consider a circle with center O.

Consider a point P 25 cm away from the center such that OP = 25 cm

A tangent PQ is drawn at point Q on the circle from point P such that PQ = 24 cm

To Find : Length of radius OQ = ?

Now, OQ ⏊ PQ

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

∴ In right - angled △POQ,

By Pythagoras Theorem,

[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]

(OP)² = (OQ)² + (PQ)²

(25)² = (OQ)² + (24)²

625 = (OQ)² + 576

(OQ)² = 49

OQ = 7 cm

Given: Two concentric circles (say C₁ and C₂) with common center as O and radius r₁ = 6.5 cm and r₂ = 2.5 cm respectively.

To Find: Length of the chord of the larger circle which touches the circle C₂. i.e. Length of AB.

As AB is tangent to circle C₂ and we know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"

So, we have,

OP ⏊ AB

∴ OPB is a right - angled triangle at P,

By Pythagoras Theorem in △OPB

[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]

We have,

(OP)² + (PB)² = (OB)²

r₂² + (PB)² = r₁²

(2.5)² + (PB)²= (6.5)²

6.25 + (PB)² = 42.25

(PB)² = 36

PB = 6 cm

Now, AP = PB ,

[ as perpendicular from center to chord bisects the chord and OP ⏊ AB ]

So,

AB = AP + PB = PB + PB

= 2PB = 2(6)

= 12 cm

Let AD = x cm, BE = y cm and CF = z cm

As we know that,

Tangents from an external point to a circle are equal,

In given Figure we have

AD = AF = x [Tangents from point A]

BD = BE = y [Tangents from point B]

CF = CE = z [Tangents from point C]

Now, Given: AB = 12 cm

AD + BD = 12

x + y = 12

y = 12 – x…. [1]

and BC = 8 cm

BE + EC = 8

y + z = 8

12 - x + z = 8 [From 1]

z = x – 4…. [2]

and

AC = 10 cm

AF + CF = 10

x + z = 10 [From 2]

x + x - 4 = 10

2x = 14

x = 7 cm

Putting value of x in [1] and [2]

y = 12 - 7 = 5 cm

z = 7 - 4 = 3 cm

So, we have AD = 7 cm, BE = 5 cm and CF = 3 cm

Given: PA and PB are tangents to a circle with center O

To show : A, O, B and P are concyclic i.e. they lie on a circle i.e. AOBP is a cyclic quadrilateral.

Proof:

OB ⏊ PB and OA ⏊ AP

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

∠OBP = ∠OAP = 90°

∠OBP + ∠OAP = 90 + 90 = 180°

AOBP is a cyclic quadrilateral i.e. A, O, B and P are concyclic.

[ As we know if the sum of opposite angles in a quadrilateral is 180° then quadrilateral is cyclic]

Hence Proved.

Given: Two concentric circles with common center as O

To Prove: AC = CB

Construction: Join OC, OA and OB

Proof :

OC ⏊ AB

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

In △OAC and △OCB, we have

OA = OB

[∵ radii of same circle]

OC = OC

[∵ common]

∠OCA = ∠OCB

[∵ Both 90° as OC ⏊ AB]

△OAC ≅ △OCB

[By Right Angle - Hypotenuse - Side]

AC = CB

[Corresponding parts of congruent triangles are congruent]

Hence Proved.

Given : From an external point P, two tangents, PA and PB are drawn to a circle with center O. At a point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. And PA = 14 cm

To Find : Perimeter of △PCD

As we know that, Tangents drawn from an external point to a circle are equal.

So we have

AC = CE …[1] [Tangents from point C]

ED = DB …[2] [Tangents from point D]

Now Perimeter of Triangle PCD

= PC + CD + DP

= PC + CE + ED + DP

= PC + AC + DB + DP [From 1 and 2]

= PA + PB

Now,

PA = PB = 14 cm as tangents drawn from an external point to a circle are equal

So we have

Perimeter = PA + PB = 14 + 14 = 28 cm

As we know that tangents drawn from an external point to a circle are equal ,

In the Given image we have,

AP = AR = 7 cm ….[1]

[tangents from point A]

CR = QC = 5 cm ….[2]

[tangents from point C]

BQ = PB …[3]

[tangents from point B]

Now,

AB = 10 cm [Given]

AP + PB = 10 cm

7 + PB = 10 [From 1]

PB = 3 cm

BQ = 3 cm …..[4]

[From 3]

BC = BQ + QC = 5 + 3 = 8 cm [ From 2 and 4]

Let sides AB, BC, CD, and AD touches circle at P, Q, R and S respectively.

As we know that tangents drawn from an external point to a circle are equal,

In the given image we have,

AP = AS = w (say) [Tangents from point A]

BP = BQ = x (say) [Tangents from point B]

CP = CR = y (say) [Tangents from point C]

DR = DS = z (say) [Tangents from point D]

Now,

Given,

AB = 6 cm

AP + BP = 6

w + x = 6 …[1]

BC = 7 cm

BP + CP = 7

x + y = 7 ….[2]

CD = 4 cm

CR + DR = 4

y + z = 4 ….[3]

Also,

AD = AS + DS = w + z ….[4]

Add [1] and [3] and substracting [2] from the sum we get,

w + x + y + z - (x + y) = 6 + 4 - 7

w + z = 3 cm ; From [4]

AD = 3 cm

As we know that tangents drawn from an external point to a circle are equal,

BR = BP [ Tangents from point B] [1]

QC = CP [ Tangents from point C] [2]

AR = AQ [ Tangents from point A] [3]

As ABC is an isosceles triangle,

AB = BC [Given] [4]

Now substract [3] from [4]

AB - AR = BC - AQ

BR = QC

BP = CP [ From 1 and 2]

∴ P bisects BC

Hence Proved.

In given Figure,

OA ⏊ AP

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

∴ In right - angled △OAP,

By Pythagoras Theorem

[i.e. (hypotenuse)² = (perpendicular)² + (base)²]

(OP)² = (OA)² + (PA)²

Given, PA = 10 cm and OA = radius of outer circle = 6 cm

(OP)² = (6)² + (100)²

(OP)² = 36 + 100 = 136 [1]

Also,

OB ⏊ BP

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

∴ In right - angled △OBP,

By Pythagoras Theorem

[i.e. (hypotenuse)² = (perpendicular)² + (base)²]

(OP)² = (OB)² + (PB)²

Now, OB = radius of inner circle = 4 cm

And from [2]

(OP)² = 136

136 = (4)² + (PB)²

(PB)² = 136 - 16 = 120

PB = 10.9 cm

Given : △ABC that is drawn to circumscribe a circle with radius r = 3 cm and BD = 6 cm DC = 9cm

Also, area(△ABC) = 54 cm²

To Find : AB and AC

Now,

As we know tangents drawn from an external point to a circle are equal.

Then,

FB = BD = 6 cm [Tangents from same external point B]

DC = EC = 9 cm [Tangents from same external point C]

AF = EA = x (let) [Tangents from same external point A]

Using the above data, we get

AB = AF + FB = x + 6 cm

AC = AE + EC = x + 9 cm

BC = BD + DC = 6 + 9 = 15 cm

Now we have heron's formula for area of triangles if its three sides a, b and c are given

Where,

⇒

So for △ABC

a = AB = x + 6

b = AC = x + 9

c = BC = 15 cm

⇒

And

Squaring both sides, we get,

54(54) = 54x(x + 15)

x² + 15x - 54 = 0

x² + 18x - 3x - 54 = 0

x(x + 18) - 3(x + 18) = 0

(x - 3)(x + 18) = 0

x = 3 or - 18

but x = - 18 is not possible as length can't be negative.

So

AB = x + 6 = 3 + 6 = 9 cm

AC = x + 9 = 3 + 9 = 12 cm

Given : A circle with center O and radius 3 cm and PQ is a chord of length 4.8 cm. The tangents at P and Q intersect at point T

To Find : Length of TP

Construction : Join OQ

Now in △OPT and △OQT

OP = OQ [radii of same circle]

PT = PQ

[tangents drawn from an external point to a circle are equal]

OT = OT [Common]

△OPT ≅ △OQT [By Side - Side - Side Criterion]

∠POT = ∠OQT

[Corresponding parts of congruent triangles are congruent]

or ∠POR = ∠OQR

Now in △OPR and △OQR

OP = OQ [radii of same circle]

OR = OR [Common]

∠POR = ∠OQR [Proved Above]

△OPR ≅ △OQT [By Side - Angle - Side Criterion]

∠ORP = ∠ORQ

[Corresponding parts of congruent triangles are congruent]

Now,

∠ORP + ∠ORQ = 180° [Linear Pair]

∠ORP + ∠ORP = 180°

∠ORP = 90°

⇒ OR ⏊ PQ

⇒ RT ⏊ PQ

As OR ⏊ PQ and Perpendicular from center to a chord bisects the chord we have

∴ In right - angled △OPR,

By Pythagoras Theorem

[i.e. (hypotenuse)² = (perpendicular)² + (base)²]

(OP)² = (OR)² + (PR)²

(3)² = (OR)² + (2.4)²

9 = (OR)² + 5.76

(OR)²= 3.24

OR = 1.8 cm

Now,

In right angled △TPR,

By Pythagoras Theorem

(PT)² = (PR)² + (TR)² …[1]

Also, OP ⏊ OT

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

In right angled △OPT, By Pythagoras Theorem

(PT)² + (OP)² = (OT)²

(PR)² + (TR)² + (OP)²= (TR + OR)² …[From 1]

(2.4)² + (TR)² + (3)² = (TR + 1.8)²

4.76 + (TR)² + 9 = (TR)² + 2(1.8)TR + (1.8)²

13.76 = 3.6TR + 3.24

3.6TR = 10.52

TR = 2.9 cm [Appx]

Using this in [1]

PT² = (2.4)² + (2.9)²

PT² = 4.76 + 8.41

PT² = 13.17

PT = 3.63 cm [Appx]

Given: A circle with center O and AB and CD are two parallel tangents at points P and Q on the circle.

To Prove: PQ passes through O

Construction: Draw a line EF parallel to AB and CD and passing through O

Proof :

∠OPB = 90°

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

Now, AB || EF

∠OPB + ∠POF = 180°

90° + ∠POF = 180°

∠POF = 90° …[1]

Also,

∠OQD = 90°

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

Now, CD || EF

∠OQD + ∠QOF = 180°

90° + ∠QOF = 180°

∠QOF = 90° [2]

Now From [1] and [2]

∠POF + ∠QOF = 90 + 90 = 180°

So, By converse of linear pair POQ is a straight Line

i.e. O lies on PQ

Hence Proved.

In quadrilateral POQB

∠OPB = 90°

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

∠OQB = 90°

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

∠PQB = 90° [Given]

By angle sum property of quadrilateral PQOB

∠OPB + ∠OQB + ∠PBQ + ∠POQ = 360°

90° + 90° + 90° + ∠POQ = 360°

∠POQ = 90°

As all angles of this quadrilaterals are 90° The quadrilateral is a rectangle

Also, OP = OQ = r

i.e. adjacent sides are equal, and we know that a rectangle with adjacent sides equal is a square

∴ POQB is a square

And OP = PB = BQ = OQ = r [1]

Now,

As we know that tangents drawn from an external point to a circle are equal

In given figure, We have

DS = DR = 5 cm

[Tangents from point D and DS = 5 cm is given]

AD = 23 cm [Given]

AR + DR = 23

AR + 5 = 23

AR = 18 cm

Now,

AR = AQ = 18 cm

[Tangents from point A]

AB = 29 cm [Given]

AQ + QB = 29

18 + QB = 29

QB = 11 cm

From [1]

QB = r = 11 cm

Hence Radius of circle is 11 cm.

In Given Figure, we have a circle with center O let the radius of circle be r.

Construction : Join OP

Now, In △APB

∠ABP = 30°

∠APB = 90°

[Angle in a semicircle is a right angle]

By angle sum Property of triangle,

∠ABP + ∠APB + ∠PAB = 180

30° + 90° + ∠PAB = 180

∠PAB = 60°

OP = OA = r [radii]

∠PAB = ∠OPA = 60°

[Angles opposite to equal sides are equal]

By angle sum Property of triangle

∠OPA + ∠OAP + ∠AOP = 180°

60° + ∠PAB + ∠AOP = 180

60 + 60 + ∠AOP = 180

∠AOP = 60°

As all angles of △OPA equals to 60°, △OPA is an equilateral triangle

So, we have, OP = OA = PA = r [1]

∠OPT = 90°

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

∠OPA + ∠APT = 90

60 + ∠APT = 90

∠APT = 30°

Also,

∠PAB + ∠PAT = 180° [Linear pair]

60° + ∠PAT = 180°

∠PAT = 120°

In △APT

∠APT + ∠PAT + ∠PTA = 180°

30° + 120° + ∠PTA = 180°

∠PTA = 30°

So,

We have

∠APT = ∠PTA = 30°

AT = PA

[Sides opposite to equal angles are equal]

AT = r [From 1] [2]

Now,

AB = OA + OB = r + r = 2r [3]

From [2] and [3]

AB : AT = 2r : r = 2 : 1

Hence Proved !

Let sides AB, BC, CD, and AD touches circle at P, Q, R and S respectively.

As we know that tangents drawn from an external point to a circle are equal ,

In the given image we have,

AP = AS = w (say) [Tangents from point A]

BP = BQ = x (say) [Tangents from point B]

CP = CR = y (say) [Tangents from point C]

DR = DS = z (say) [Tangents from point D]

Now,

Given,

AB = 6 cm

AP + BP = 6

w + x = 6 [1]

BC = 9 cm

BP + CP = 9

x + y = 9 [2]

CD = 8 cm

CR + DR = 8

y + z = 8 [3]

Also,

AD = AS + DS = w + z [4]

Add [1] and [3] and substracting [2] from the sum we get,

w + x + y + z - (x + y) = 6 + 8 - 9

w + z = 5 cm

From [4]

AD = 5 cm

In the given figure, PA and PB are two tangents from common point P

∴ PA = PB

[Tangents drawn from an external point are equal]

∠PBA = ∠PAB

[Angles opposite to equal angles are equal] [1]

By angle sum property of triangle in △APB

∠APB + ∠PBA + ∠PAB = 180°

50° + ∠PAB + ∠PAB = 180° [From 1]

2∠PAB = 130°

∠PAB = 65° [2]

Now,

∠OAP = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OAB + ∠PAB = 90°

∠OAB + 65° = 90° [From 2]

∠OAB = 25°

Given: In the figure, PT and PQ are two tangents and ∠TPQ = 70°

To Find: ∠TRQ

Construction: Join OT and OQ

In quadrilateral OTPQ

∠OTP = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OQP = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠TPQ = 70° [Common]

By Angle sum of Quadrilaterals,

In quadrilateral OTPQ we have

∠OTP + ∠OQP + ∠TPQ + ∠TOQ = 360°

90° + 90° + 70° + ∠TOQ = 360°

250° + ∠TOQ = 360

∠TQO = 110°

Now,

As we Know the angle subtended by an arc at the center is double the angle subtended by it at any

point on the remaining part of the circle.

∴ we have

∠TOQ = 2∠TRQ

110° = 2 ∠TRQ

∠TRQ = 55°

Given: AB and CD are two tangents to two circles which intersects at E .

To Prove: AB = CD

Proof:

As

AE = CE …[1]

[Tangents drawn from an external point to a circle are equal]

And

EB = ED …[2]

[Tangents drawn from an external point to a circle are equal]

Adding [1] and [2]

AE + EB = CE + ED

AB = CD

Hence Proved.

Given: PT is a tangent to a circle with center O and PQ is a chord of the circle such that ∠QPT = 70°

To Find: ∠POQ = ?

Now,

∠OPT = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OPQ + ∠QPT = 90°

∠OPQ + 70° = 90°

∠OPQ = 20°

Also,

OP = OQ [Radii of same circle]

∠OQP = ∠OPQ = 20°

[Angles opposite to equal sides are equal]

In △OPQ By Angle sum property of triangles,

∠OPQ + ∠OQP + ∠POQ = 180°

20° + 20° + ∠POQ = 180°

∠POQ = 140°

Given: △ABC that is drawn to circumscribe a circle with radius r = 2 cm and BD = 4 cm DC = 3cm

Also, area(△ABC) = 21 cm²

To Find: AB and AC

Now,

As we know tangents drawn from an external point to a circle are equal.

Then,

FB = BD = 4 cm [Tangents from same external point B]

DC = EC = 3 cm [Tangents from same external point C]

AF = EA = x (let) [Tangents from same external point A]

Using the above data, we get

AB = AF + FB = x + 4 cm

AC = AE + EC = x + 3 cm

BC = BD + DC = 4 + 3 = 7 cm

Now we have heron's formula for area of triangles if its three sides a, b and c are given

Where,

So, for △ABC

a = AB = x + 4

b = AC = x + 3

c = BC = 7 cm

⇒

And

Squaring both sides,

21(21) = 12x(x + 7)

12x² + 84x - 441 = 0

4x² + 28x - 147 = 0

As we know roots of a quadratic equation in the form ax² + bx + c = 0 are,

So roots of this equation are,

but x = - 10.5 is not possible as length can't be negative.

So

AB = x + 4 = 3.5 + 4 = 7.5 cm

AC = x + 3 = 3.5 + 3 = 6.5 cm

Given : Two concentric circles (say C₁ and C₂) with common center as O and radius r₁ = 5 cm and r₂ = 3 cm respectively.

To Find : Length of the chord of the larger circle which touches the circle C₂. i.e. Length of AB.

As AB is tangent to circle C₂ and,

We know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"

So, we have,

OP ⏊ AB

∴ OPB is a right - angled triangle at P,

By Pythagoras Theorem in △OPB

[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]

We have,

(OP)² + (PB)² = (OB)²

r₂² + (PB)² = r₁²

(3)² + (PB)²= (5)²

9 + (PB)² = 25

(PB)² = 16

PB = 4 cm

Now, AP = PB,

[ as perpendicular from center to chord bisects the chord and OP ⏊ AB ]

So,

AB = AP + PB = PB + PB

= 2PB = 2(4) = 8 cm

Let us consider a circle with center O and XY be a tangent

To prove : Perpendicular at the point of contact of the tangent to a circle passes through the center i.e. the radius OP ⏊ XY

Proof :

Take a point Q on XY other than P and join OQ .

The point Q must lie outside the circle. (because if Q lies inside the circle, XY

will become a secant and not a tangent to the circle).

∴ OQ is longer than the radius OP of the circle. That is,

OQ > OP.

Since this happens for every point on the line XY except the point P, OP is the

shortest of all the distances of the point O to the points of XY.

So OP is perpendicular to XY.

[As Out of all the line segments, drawn from a point to points of a line not passing through the point, the smallest is the perpendicular to the line.]

Given : In the figure ,

Two tangents RQ and RP are drawn from an external point R to the circle with center O and ∠PRQ = 120°

To Prove: OR = PR + RQ

Construction: Join OP and OQ

Proof :

In △△OPR and △OQR

OP = OQ [radii of same circle]

OR = OR [Common]

PR = PQ …[1]

[Tangents drawn from an external point are equal]

△OPR ≅ △OQR

[By Side - Side - Side Criterion]

∠ORP = ∠ORQ

[Corresponding parts of congruent triangles are congruent]

Also,

∠PRQ = 120°

∠ORP + ∠ORQ = 120°

∠ORP + ∠ORP = 120°

2∠ORP = 120°

∠ORP = 60°

Also, OP ⏊ PR

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, In right angled triangle OPR,

∴ OR = 2PR

OR = PR + PR

OR = PR + RQ [From 1]

Hence Proved.

Let AD = x cm, BE = y cm and CF = z cm

As we know that,

Tangents from an external point to a circle are equal,

In given Figure we have

AD = AF = x

[Tangents from point A]

BD = BE = y

[Tangents from point B]6CF = CE = z [Tangents from point C]

Now, Given: AB = 14 cm

AD + BD = 14

x + y = 14

y = 14 - x … [1]

and BC = 8 cm

BE + EC = 8

y + z = 8

14 - x + z = 8 … [From 1]

z = x - 6 [2]

and

CA = 12 cm

AF + CF = 12

x + z = 12 [From 2]

x + x - 6 = 12

2x = 18

x = 9 cm

Putting value of x in [1] and [2]

y = 14 - 9 = 5 cm

z = 9 - 6 = 3 cm

So, we have AD = 9 cm, BE = 5 cm and CF = 3 cm

Given : PA and PB are tangents to a circle with center O

To show : AOBP is a cyclic quadrilateral.

Proof :

OB ⏊ PB and OA ⏊ AP

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

∠OBP = ∠OAP = 90°

∠OBP + ∠OAP = 90 + 90 = 180°

AOBP is a cyclic quadrilateral

[ As we know if the sum of opposite angles in a quadrilateral is 180° then quadrilateral is cyclic ]

Hence Proved.

Let us consider circles C₁ and C₂ with common center as O. Let AB be a tangent to circle C₁ at point P and chord in circle C₂. Join OB

In △OPB

OP ⏊ AB

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∴ OPB is a right - angled triangle at P,

By Pythagoras Theorem,

[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)²]

(OB)² = (OP)² + (PB)²

Now, 2PB = AB

[As we have proved above that OP ⏊ AB and Perpendicular drawn from center to a chord bisects the chord]

2PB = 8 cm

PB = 4 cm

(OB)² = (5)² + (4)²

[As OP = 5 cm, radius of inner circle]

(OB)² = 41

⇒ OB = √41 cm

Given : , PQ is a chord of a circle with center 0 and PT is a tangent and ∠QPT = 60°.

To Find : ∠PRQ

∠OPT = 90°

∠OPQ + ∠QPT = 90°

∠OPQ + 60° = 90°

∠OPQ = 30° … [1]

Also.

OP = OQ [radii of same circle]

∠OQP = ∠OPQ [Angles opposite to equal sides are equal]

From [1], ∠OQP = ∠OQP = 30°

In △OPQ , By angle sum property

∠OQP + ∠OPQ + ∠POQ = 180°

30° + 30° + ∠POQ = 180°

∠POQ = 120°

As we know, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

So, we have

2∠PRQ = reflex ∠POQ

2∠PRQ = 360° - ∠POQ

2∠PRQ = 360° - 120° = 240°

∠PRQ = 120°

In the given figure, PA and PB are two tangents from common point P

∴ PA = PB

[∵ Tangents drawn from an external point are equal]

∠PBA = ∠PAB

[∵ Angles opposite to equal angles are equal] …[1]

By angle sum property of triangle in △APB

∠APB + ∠PBA + ∠PAB = 180°

60° + ∠PAB + ∠PAB = 180° [From 1]

2∠PAB = 120°

∠PAB = 60° …[2]

Now,

∠OAP = 90° [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OAB + ∠PAB = 90°

∠OAB + 60° = 90° [From 2]

∠OAB = 30°

The maximum number of tangents that can be drawn from an external point to a circle is two and they are equal in length.

As SQ is diameter and OQ is radius in the given circle,

∴ 2OQ = SQ [As 2 × radius) = diameter]

2OQ = 6 cm

OQ = 3 cm

Now, QR is tangent

∴ OQ ⏊ QR

In right - angled △OQR,

By Pythagoras Theorem,

[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)² ]

(QR)² + (OQ)² = (OR)²

(4)² + (3)² = (OR)²

16 + 9 = (OR)²

(OR)² = 25

OR = 5 cm

We have given, PT is a tangent drawn at point T on the circle.

∴ OT ⏊ TP

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, In △OTP we have,

By Pythagoras Theorem,

[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)²]

(OP)² = (OT)² + (PT)²

(OP)² = (7)² + (24)²

(OP)² = 49 + 576

(OP)² = 625

⇒ OP = 25 cm

As all diameters of a circle passes through center O it is not possible to have two parallel diameters in a circle.

Let us consider a circle with center O and AB be any chord that subtends 90° angle at its center.

Now, In △OAB

OA = OB = 10 cm

And as ∠AOB = 90° ,

By Pythagoras Theorem,

[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)²]

(OA)² + (OB)² = (AB)²

(10)²+ (10)² = (AB)²

100 + 100 = (AB)²

⇒ AB = √200 = 10√2

So, Correct option is C .

We have given, PT is a tangent drawn at point T on the circle.

∴ OT ⏊ TP

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, In △OTP we have,

By Pythagoras Theorem,

[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)²]

(OP)² = (OT)² + (PT)²

(10)² = (6)² + (PT)²

(PT)² = 100 - 36

(PT)² = 64

⇒ PT = 8 cm

We have given, PT is a tangent drawn at point T on the circle and OP = 26 cm and PT = 24 cm

Join OT

∴ OT ⏊ TP

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, In △OTP we have,

By Pythagoras Theorem,

[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)²]

(OP)² = (OT)² + (PT)²

(26)² = (OT)² + (24)²

(OT)² = 676 - 576

(OT)² = 100

OT = 10 cm

Hence, radius of circle is 10 cm.

Let us consider a circle with center O and PQ is a tangent

on the circle, Joined OP and OQ

But OPQ is an isosceles triangle, ∴ OP = PQ

∠OQP = ∠POQ

[Angles opposite to equal sides are equal]

In △OQP

∠OQP + ∠OPQ + ∠POQ = 180°

[Angle sum property of triangle]

∠OQP + 90° + ∠OPQ = 180°

2 ∠OPQ = 90°

∠OPQ = 45°

As AB and AC are tangents to given circle,

We have,

OB ⏊ AB and OC ⏊ AC

[∵ Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠OBA = ∠OCA = 90°

In quadrilateral AOBC, By angle sum property of quadrilateral, we have,

∠OBA + ∠OCA + ∠BOC + ∠BAC = 360°

90° + 90° + ∠BOC + 40° = 360°

∠BOC = 140°

Let us consider a circle with center O and AB be a chord such that ∠AOB = 60°

AP and BP are two intersecting tangents at point P at point A and B respectively on the circle.

To find : Angle between tangents, i.e. ∠APB

As AP and BP are tangents to given circle,

We have,

OA ⏊ AP and OB ⏊ BP [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠OAP = ∠OBP = 90°

In quadrilateral AOBP, By angle sum property of quadrilateral, we have

∠OAP + ∠OBP + ∠APB + ∠AOB = 360°

90° + 90° + ∠APB + 60° = 360°

∠APB = 120°

Given: Two concentric circles (say C₁ and C₂) with common center as O and radius r₁ = 6 cm(inner circle) and r₂ = 10 cm (outer circle) respectively.

To Find : Length of the chord AB.

As AB is tangent to circle C₁ and we know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"

So, we have,

OP ⏊ AB

∴ OPB is a right - angled triangle at P,

By Pythagoras Theorem in △OPB

[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]

We have,

(OP)² + (PA)² = (OA)²

r₁² + (PA)² = r₂²

(6)² + (PA)²= (10)²

36 + (PA)² = 100

(PA)² = 64

PA = 8 cm

Now, PA = PB ,

[ as perpendicular from center to chord bisects the chord and OP ⏊ AB]

So,

AB = PA + PB = PA + PA = 2PA = 2(8) = 16 cm

As AB is tangent to the circle at point B

OB ⏊ AB

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

In right angled triangle AOB,

By Pythagoras Theorem,

[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)² ]

(OA)² = (OB)² + (AB)²

(17)² = (8)² + (AB)²

[As OA = 17 cm is given and OB is radius]

289 = 64 + (AB)²

(AB)² = 225

AB = 15 cm

Now, AB = AC [Tangents drawn from an external point are equal]

∴ AC = 15 cm

In △ABC

∠ABC = 90°

[Angle in a semicircle is a right angle]

∠ACB = 50° [Given]

By angle sum Property of triangle,

∠ACB + ∠ABC + ∠CAB = 180°

90° + 50° + ∠CAB = 180°

∠CAB = 40°

Now,

∠CAT = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠CAB + ∠BAT = 90°

40° + ∠BAT = 90°

∠BAT = 50°

In △OPQ

∠POQ = 70° [Given]

OP = OQ [radii of same circle]

∠OQP = ∠OPQ [Angles opposite to equal sides are equal]

By angle sum Property of triangle,

∠POQ + ∠OQP + ∠OPQ = 180°

70° + ∠OPQ + ∠OPQ = 180°

2 ∠OPQ = 110°

∠OPQ = 55°

Now,

∠OPT = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OPQ + ∠TPQ = 90°

55° + ∠TPQ = 90°

∠TPQ = 35°

Given: AT is a tangent to the circle with center O such that OT = 4 cm and ∠OTA = 30°.

To find: The value of AT.

Solution:




In △ OAT ,

OA ⏊ AT [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∴OAT is a right - angled triangle at A and

AT = 2√3 cm

As AP and BP are tangents to given circle,

We have,

OA ⏊ AP and OB ⏊ BP

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠OAP = ∠OBP = 90°

In quadrilateral AOBP,

By angle sum property of quadrilateral, we have

∠OAP + ∠OBP + ∠AOB + ∠APB = 360°

90° + 90° + 110° + ∠APB = 360°

∠APB = 70°

As we know,

Tangents drawn from an external point are equal, We have

AF = AE = 4 cm

[Tangents from common point A]

BF = BD = 3 cm

[Tangents from common point B]

CE = CD = x (say)

[Tangents from common point C]

Now,

AC = AE + CE

11 = 4 + x

x = 7 cm [1]

and, BC = BD + BC

BC = 3 + x = 3 + 7 = 10 cm

In the given figure PT is a tangent to circle ∴ we have

∠OPT = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OPQ + ∠QPT = 90°

∠OPQ + 50° = 90°

∠OPQ = 40°

Now, In △POQ

OP = OQ

∠PQO = ∠QPO = 40°

[Angles opposite to equal sides are equal]

Now,

∠ PQO + ∠QPO + ∠ POQ = 180°

[By angle sum property of triangle]

40° + 40° + ∠POQ = 180°

∠POQ = 100°

In the given figure, PA and PB are two tangents from common point P

∴ PA = PB

[Tangents drawn from an external point are equal]

∠PBA = ∠PAB…[1]

[Angles opposite to equal angles are equal]

By angle sum property of triangle in △APB

∠APB + ∠PBA + ∠PAB = 180°

60° + ∠PAB + ∠PAB = 180° [From 1]

2∠PAB = 120°

∠PAB = 60°…[2]

Now,

∠OAP = 90° [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OAB + ∠PAB = 90°

∠OAB + 60° = 90° [From 2]

∠OAB = 30°

Let us consider a circle with center O and AP and BP are two tangents such that angle of inclination i.e. ∠APB = 60°

Joined OA, OB and OP.

To Find : Length of tangents

Now,

PA = PB [Tangents drawn from an external point are equal] [1]

In △AOP and △BOP

PA = PB [By 1]

OP = OP [Common]

OA = OB [radii of same circle]

△AOP ≅△BOP

[By Side - Side - Side Criterion]

∠OPA = ∠OPB

[Corresponding parts of congruent triangles are congruent]

Now,

∠APB = 60° [Given]

∠OPA + ∠OPB = 60°

∠OPA + ∠OPA = 60°

2 ∠OPA = 60°

∠OPA = 30°

In △AOP

OA ⏊ PA

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact

∴ AOP is a right - angled triangle.

So, we have

⟹PA = 3√3 cm

From [1]

PA = PB = 4 cm

i.e. length of each tangent is 3√3 cm

In Given Figure,

PQ = PR…[1]

[Tangents drawn from an external point are equal]

In △AOP and △BOP

PQ = PR [By 1]

AP = AP [Common]

AQ = AR [radii of same circle]

△AQP ≅△ARP [By Side - Side - Side Criterion]

∠QPA = ∠RPA

[Corresponding parts of congruent triangles are congruent]

Now,

∠QPA + ∠RPA = ∠QPR

∠QPA + ∠QPA = ∠QPR

2 ∠QPA = ∠QPR

∠QPR = 2(27) = 54°

As PQ and PQ are tangents to given circle,

We have,

AQ ⏊ PQ and AR ⏊ PR

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠AQP = ∠ARP = 90°

In quadrilateral AQRP, By angle sum property of quadrilateral, we have

∠AQP + ∠ARP + ∠QAR + ∠QPR = 360°

90° + 90° + ∠QAR + 54° = 360°

∠QAR = 126°

Join AC, BC and CP

To Find: Length of tangents

Now,

PA = PB…[1]

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

In △ACP and △BCP

PA = PB [By 1]

CP = CP [Common]

CA = CB [radii of same circle]

△ACP ≅△BCP [By Side - Side - Side Criterion]

∠CPA = ∠CPB

[Corresponding parts of congruent triangles are congruent]

Now,

∠APB = 90°

[Given that PA ⏊ PB]

∠CPA + ∠CPB = 90°

∠CPA + ∠CPA = 90°

2 ∠CPA = 90°

∠CPA = 45°

In △ACP

CA ⏊ PA [Tangents drawn at a point on circle is perpendicular to the radius through point of contact

∴ ACP is a right - angled triangle.

So, we have

⟹PA = 4 cm

From [1]

PA = PB = 4 cm

i.e. length of each tangent is 4 cm

In Given Figure,

PA = PB…[1]

[Tangents drawn from an external point are equal]

In △AOP and △BOP

PA = PB [By 1]

OP = OP [Common]

OA = OB

[radii of same circle]

△AOP ≅△BOP

[By Side - Side - Side Criterion]

∠OPA = ∠OPB

[Corresponding parts of congruent triangles are congruent]

Now,

∠APB = 80° [Given]

∠OPA + ∠OPB = 80°

∠OPA + ∠OPA = 80°

2 ∠OPA = 80°

∠OPA = 40°

In △AOP,

∠OAP = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

And

∠OAP + ∠OPA + ∠AOP = 180°

90° + 40° + ∠AOP = 180°

∠AOP = 50°

In the given Figure, Join OP

Now, OP ⏊ AB

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∴∠OPA = 90°

∠OPQ + ∠APQ = 90°

∠OPQ + 58° = 90°

[Given, ∠APQ = 58°]

∠OPQ = 32°

In △OPQ

OP = OQ

[Radii of same circle]

∠OQP = ∠OPQ

[Angles opposite to equal sides are equal]

∠PQB = 32°

[As ∠OQP = ∠PQB ]

In given Figure, Join OP

In △OPC,

OP = OC [Radii of same circle]

∠OCP = ∠OPC

[Angles opposite to equal sides are equal]

∠ACP = ∠OPC

[As ∠OCP = ∠ACP] …[1]

Now,

∠OPB = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OPC + ∠CPB = 90°

∠ACP + ∠CPB = 90° [By 1]

So,

∠CPB + ∠ACP = 90°

In the given Figure, Join OA

Now,

OA ⏊ PQ

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OAP = ∠OAQ = 90° [1]

∠OAB + ∠PAB = 90°

∠OAB + 67° = 90°

∠OAB = 23°

Now,

∠BAC = 90°

[Angle in a semicircle is a right angle]

∠OAB + ∠OAC = 90°

23° + ∠OAC = 90°

∠OAC = 67°

∠OAQ = 90° [From 1]

∠OAC + ∠CAQ = 90°

67° + ∠CAQ = 90°

∠CAQ = 23° [2]

Now,

OA = OC

[radii of same circle]

∠OCA = ∠OAC

[Angles opposite to equal sides are equal]

∠OCA = 67°

∠OCA + ∠ACQ = 180° [Linear Pair]

67° + ∠ACQ = 180°

∠ACQ = 113° [3]

Now, In △ACQ By Angle Sum Property of triangle

∠ACQ + ∠CAQ + ∠AQC = 180°

113° + 23° + ∠AQC = 180° [By 2 and 3]

∠AQC = 44°

In Given Figure,

PQ = PR…[1]

[Tangents drawn from an external point are equal]

In △QOP and △ROP

PQ = PR [By 1]

OP = OP [Common]

OQ = OR [radii of same circle]

△QOP ≅△ROP

[By Side - Side - Side Criterion]

area(Δ QOP) = area(Δ ROP)

[Congruent triangles have equal areas]

area(PQOR) = area(Δ QOP) + area(Δ ROP)

area(PQOR) = area(Δ QOP) + area(Δ QOP) = 2[area(Δ QOP)]

Now,

OQ ⏊PQ

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, QOP is a right - angled triangle at Q with OQ as base and PQ as height.

In △QOP,

By Pythagoras Theorem in △OPB

[i.e. (hypotenuse)² = (perpendicular)² + (base)²]

(OQ)² + (PQ)² = (OP)²

(5)² + (PQ)² = (13)²

25 + (PQ)² = 169

(PQ)² = 144

PQ = 12 cm

Area(ΔQOP) = 1/2 × Base × Height

= 1/2 × OQ × PQ

= 1/2 × 5 × 12

= 30 cm²

So,

Area(PQOR) = 2(30) = 60 cm²

In given figure, as PR is a tangent

OQ ⏊ PR

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

⟹LQ ⏊ PR

⟹LQ ⏊ AB

[As, AB || PR]

AL = LB

[Perpendicular from center to the chord bisects the chord]

Now,

∠LQR = 90°

∠LQB + ∠BQR = 90°

∠LQB + 70° = 90°

∠LQB = 20°…[1]

In △AQL and △BQL

∠ALQ = ∠BLQ [Both 90° as LQ ⏊ AB]

AL = LB [Proved above]

QL = QL [Common]

△AQL ≅ △BQL

[Side - Angle - Side Criterion]

∠LQA = ∠LQB

[Corresponding parts of congruent triangles are congruent]

∠AQB = ∠LQA + ∠LQB = ∠LQB + ∠LQB

= 2∠LQB = 2(20) = 40° [By 1]

Let us consider a circle with center O and TP be a tangent at point A on the circle, Joined OT and OP

Given Length of tangent, TP = 10 cm, and OT = 5 cm [radius]

To Find : Distance of center O from P i.e. OP

Now,

OP ⏊ TP

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So OPT is a right - angled triangle,

By Pythagoras Theorem in ΔOPB

[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]

(OT)² + (TP)² = (OP)²

(OP)² = (5)² + (10)²

(OP)² = 25 + 100 = 125

OP = √125 cm

In △BOP

OB = OP [radii of same circle]

∠OPB = ∠PBO

[Angles opposite to equal sides are equal]

As, ∠PBO = 30°

∠OPB = 30°

Now,

∠OPT = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠BPT = ∠OPB + ∠OPT = 30° + 90° = 120°

Now, In ΔBPT

∠BPT + ∠PBO + ∠PTB = 180°

120° + 30° + ∠PTB = 180°

∠PTB = 30°

∠PTA = ∠PTB = 30°

Given : In the given figure, a circle touches the side DF of ΔEDF at H and touches ED and EF produced at K and M respectively and EK = 9 cm

To Find : Perimeter of △EDF

As we know that, Tangents drawn from an external point to a circle are equal.

So, we have

KD = DH …[1]

[Tangents from point D]

HF = FM …[2]

[Tangents from point F]

Now Perimeter of Triangle PCD

= ED + DF + EF

= ED + DH + HF + EF

= ED + KD + FM + EF [From 1 and 2]

= EK + EM

Now,

EK = EM = 9 cm as tangents drawn from an external point to a circle are equal

So, we have

Perimeter = EK + EM = 9 + 9 = 18 cm

Let us consider a circle with center O and PA and PB are two tangents from point P, given that angle of inclination i.e. ∠APB = 45°

As PA and PB are tangents to given circle,

We have,

OA ⏊ PA and OB ⏊ PB [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠OAP = ∠OBP = 90°

In quadrilateral AQRP,

By angle sum property of quadrilateral, we have

∠OAP + ∠OBP + ∠ABP + ∠AOB = 360°

90° + 90° + 45° + ∠AOB = 360°

∠AOB = 135°

As PL and PM are tangents to given circle,

We have,

OR ⏊ PM and OQ ⏊ PL

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠ORM = ∠OQL = 90°

∠ORM = ∠ORS + ∠SRM

90° = ∠ORS + 60°

∠ORS = 30°

And ∠OQL = ∠OQS + ∠SQL

90° = ∠OQS + 50°

∠OQS = 40°

Now, In △SOR

OS = OQ [radii of same circle]

∠ORS = ∠OSR

[Angles opposite to equal sides are equal]

∠OSR = 30°

[as ∠ORS = 30°]

Now, In △SOR

OS = SQ [radii of same circle]

∠OQS = ∠OSQ

[Angles opposite to equal sides are equal]

∠OSQ = 40° [as ∠OQS = 40°]

As,

∠QSR = ∠OSR + ∠OSQ

∠QSR = 30° + 40° = 70°

Given : △PQR that is drawn to circumscribe a circle with radius r = 6 cm and QT = 12 cm QR = 9cm

Also, area(△PQR) = 189 cm²

Let tangents PR and PQ touch the circle at X and Y respectively.

To Find : PQ and QR

Now,

As we know tangents drawn from an external point to a circle are equal.

Then,

QT = QY = 12 cm

[Tangents from same external point B]

TR = RX = 9 cm

[Tangents from same external point C]

PX = PY = x (let)

[Tangents from same external point A]

Using the above data we get

PQ = PY + QT = x + 12 cm

PR = PC + RX = x + 9 cm

QR = QT + TR = 12 + 9 = 21 cm

Now we have heron's formula for area of triangles if its three sides a, b and c are given

Where,

So for △PQR

a = PQ = x + 12

b = PR = x + 9

c = QR = 21 cm

And

Squaring both side

189(189) = 108(x + 21)

7(189) = 4(x + 21)

4x² + 84x - 1323 = 0

As we know roots of a quadratic equation in the form ax² + bx + c = 0 are,

So, roots of this equation are,

but x = - 31.5 is not possible as length can't be negative.

So

PQ = x + 12 = 10.5 + 12 = 22.5 cm

Let the bigger circle be C1 and Smaller be C2,

Now,

PQ and PT are two tangents to circle C1,

∴ PT = QP

[Tangents drawn from an external point are equal]

QP = 3.8 cm

[ As PT = 3.8 cm is given]

Also,

PR and PT are two tangents to circle C2,

∴ PT = PR

[Tangents drawn from an external point are equal]

PR = 3.8 cm

[ As PT = 3.8 cm is given]

QR = QP + PR = 3.8 + 3.8 = 7.6 cm

As we know Tangents drawn from an external point are equal]

In the given Figure, we have

AP = AQ = 5 cm

[Tangents from point A] [AP = 5 cm is given]

BQ = BR = x(say)

[Tangents from point B]

CR = CS = 3 cm [∵ CS = 3 cm is given]

[Tangents from point C]

Given,

BC = 7 cm

CR + BR = 7

3 + x = 7 cm

x = 4 cm

Now,

AB = AQ + BQ = 5 + x = 5 + 4 = 9 cm

As we know Tangents drawn from an external point are equal]

In the given Figure, we have

AP = AS = 6 cm [AP = 6 cm is given]

[∵ Tangents from point A]

BP = BQ = 5 cm [BP = 5 cm is given]

[∵ Tangents from point B]

CR = CQ = 3 cm [CQ = 3 cm is given]

[∵ Tangents from point C]

DR = DS = 4 cm ][DR = 4 cm is given]

[∵ Tangents from point D]

Now,

Perimeter of ABCD = AB + BC + CA + DA

= AP + BP + BQ + CQ + CR + DR + DS + AS

= 6 + 5 + 5 + 3 + 3 + 4 + 4 + 6 = 36 cm

In △AOB

OA = OB [radii of same circle]

∠OBA = ∠OAB [Angles opposite to equal sides are equal]

Also, By Triangle sum Property

∠AOB + ∠OBA + ∠OAB = 180°

100 + ∠OAB + ∠OAB = 180°

2 ∠OAB = 90°

∠OAB = 40°

As AT is tangent to given circle,

We have,

OA ⏊ AT

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠OAT = 90°

∠OAB + ∠BAT = 90°

40° + ∠BAT = 90°

∠BAT = 50°

Let AB, BC and AC touch the circle at points P, Q and R respectively.

As ABC is a right triangle,

By Pythagoras Theorem

[i.e. (hypotenuse)² = (perpendicular)² + (base)²]

(AC)² = (BC)² + (AB)²

(AC)² = (12)² + (5)²

(AC)² = 144 + 25 = 169

AC = 13 cm

Let O be the center of circle, Join OP, OQ and PR

Let the radius of circle be r,

We have

r = OP = OQ = OR

[radii of same circle] [1]

Now,

ar(△ABC) = ar(△AOB) + ar(△BOC) + ar(△AOC)

As we know,

Area of triangle is 1/2 × Base × Height (Altitude)

Now,

OP ⏊ AB

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∴ OP is the altitude in △AOB

OQ ⏊ BC

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∴ OQ is the altitude in △BOC

OR ⏊ AC

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∴ OR is the altitude in △AOC

So, we have

1/2 × BC × AB = (1/2 × AB × OP) + (1/2 × BC × OQ) + (1/2 × AC × OR)

12(5) = 5(r) + 12(r) + 13(r) [Using 1]

60 = 30r

r = 2 cm

In quadrilateral ORDS

∠ORD = 90°

[∵ Tangent at any point on the circle is perpendicular to the radius through point of contact]

∠OSD = 90°

[∵ Tangent at any point on the circle is perpendicular to the radius through point of contact]

∠SDR = 90° [AD ⏊ CD]

By angle sum property of quadrilateral PQOB

∠ORD + ∠OSD + ∠SDR + ∠SOR = 360°

90° + 90° + 90° + ∠SOR = 360°

∠SOR = 90°

As all angles of this quadrilaterals are 90° The quadrilateral is a rectangle

Also, OS = OR = r

i.e. adjacent sides are equal, and we know that a rectangle with adjacent sides equal is a square

∴ POQB is a square

And OS = OR = DR = DS = r = 10 cm [1]

Now,

As we know that tangents drawn from an external point to a circle are equal

In given figure, We have

CQ = CR …[2]

[∵ tangents from point C]

PB = BQ = 27 cm

[∵Tangents from point B and PB = 27 cm is given]

BC = 38 cm [Given]

BQ + CQ = 38

27 + CQ = 38

CQ = 11 cm

From [2]

CQ = CR = 11 cm

Now,

CD = CR + DR

CD = 11 + 10 = 21 cm [from 1, DR = 10 cm]

As ABC is a right triangle,

By Pythagoras Theorem

[i.e. (hypotenuse)² = (perpendicular)² + (base)²]

(AC)² = (BC)² + (AB)²

(AC)² = (6)² + (8)²

(AC)² = 36 + 64 = 100

AC = 10 cm

Now,

ar(△ABC) = ar(△AOB) + ar(△BOC) + ar(△AOC)

As we know,

Area of triangle is 1/2 × Base × Height(Altitude)

Now,

OP ⏊ AB [Given]

∴ OP is the altitude in △AOB

OQ ⏊ BC [Given]

∴ OQ is the altitude in △BOC

OR ⏊ AC [Given]

∴ OR is the altitude in △AOC

So, we have

1/2 × BC × AB = (1/2 × AB × OP) + (1/2 × BC × OQ) + (1/2 × AC × OR)

6(8) = 8(x) + 6(x) + 10(x)

[∵ OP = OQ = OR = x, Given]

48 = 24x

x = 2 cm

Let sides AB, BC, CD, and AD touches circle at P, Q, R and S respectively.

As we know that tangents drawn from an external point to a circle are equal,

So, we have,

AP = AS = w (say)

[∵ Tangents from point A]

BP = BQ = x (say)

[∵Tangents from point B]

CP = CR = y (say)

[∵Tangents from point C]

DR = DS = z (say)

[∵Tangents from point D]

Now,

Given,

AB = 6 cm

AP + BP = 6

w + x = 6 …[1]

BC = 7 cm

BP + CP = 7

x + y = 7 …[2]

CD = 4 cm

CR + DR = 4

y + z = 4 …[3]

Also,

AD = AS + DS = w + z …[4]

Add [1] and [3] and substracting [2] from the sum we get,

w + x + y + z - (x + y) = 6 + 4 - 7

w + z = 3 cm

From [4]

AD = 3 cm

.

In the given Figure

AR = AP = x(let) [Radii of same circle]

BP = BQ = y(let) [Radii of same circle]

CR = CQ = z(let) [Radii of same circle]

Now,

AB = 5 cm [Given]

AP + BP = 5

x + y = 5

y = 5 - x …[1]

BC = 7 cm [Given]

BQ + CQ = 7

y + z = 7

5 - x + z = 7 [using 1]

z = 2 + x …[2]

and

AC = 6 cm [Given]

x + z = 6

x + 2 + x = 6 [Using 2]

2x = 4

x = 2 cm

Let tangent BC touch the circle at point R

As we know tangents drawn from an external point to a circle are equal.

We have

AP = AQ

[tangents from point A]

BP = BR …[1]

[tangents from point B]

CQ = CR …[2]

[tangents from point C]

Now,

AP = AQ

⇒ AB + BP = AC + CQ

⇒ 5 + BR = 6 + CR [From 1 and 2]

⇒CR = BR - 1 …[3]

Now,

BC = 4 cm

BR + CR = 4

BR + BR - 1 = 4 [Using 3]

2BR = 5 cm

BR = 2.5 cm

BP = BR = 2.5 cm [Using 2]

AP = AB + BP = 5 + 2.5 = 7.5 cm

In given Figure,

OA ⏊ AP [Tangent at any point on the circle is perpendicular to the radius through point of contact]

∴ In right - angled △OAP,

By Pythagoras Theorem

[i.e. (hypotenuse)² = (perpendicular)² + (base)²]

(OP)² = (OA)² + (PA)²

Given, PA = 12 cm and OA = radius of outer circle = 5 cm

(OP)² = (5)² + (12)²

(OP)² = 25 + 144 = 136

OP = 13 cm …[1]

Also,

OB ⏊ BP [Tangent at any point on the circle is perpendicular to the radius through point of contact]

∴ In right - angled △OBP,

By Pythagoras Theorem

[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]

(OP)² = (OB)² + (PB)²

Now, OB = radius of inner circle = 3 cm

And, from [2] (OP) = 13 cm

(13)² = (3)² + (PB)²

(PB)² = 169 - 9 = 160

PB = 4√10 cm

A circle cannot have more than two tangents parallel, because tangents to be parallel they should be at diametrically ends and a diameter has two ends only.

A straight line can meet a circle at two points in case if it is a chord or diameter or a line intersecting the circle at two points.

If a tangent is drawn from a point inside a circle, it will intersect the circle at two points, so no tangent can be drawn from a point inside the circle.

Let us consider a circle with center O and radius 12 cm

A tangent PQ is drawn at point P such that PQ = 16 cm

To Find : Length of OQ

Now, OP ⏊ PQ [Tangent at any point on the circle is perpendicular to the radius through point of contact]

∴ In right - angled △POQ,

By Pythagoras Theorem

[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]

(OQ)² = (OP)² + (PQ)²

(OQ² = (12)² + (16)²

625 = 144 + 256

(OQ)² = 400

OQ = 20 cm

So,

Assertion is correct, and Reason is also correct.

Let PT and PQ are two tangents from external point P to a circle with center O

In △OPT and △OQT

OP = OQ

[radii of same circle]

OT = OT

[common]

PT = PQ

[Tangents drawn from an external point are equal]

△OPT ≅ △OQT

[By Side - Side - Side Criterion]

∠POT = ∠QOT

[Corresponding parts of congruent triangles are congruent]

i.e. Assertion is true

Now,

Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.

As ABCD is a parallelogram

AB = CD and BC = AD

[opposite sides of a parallelogram are equal] [1]

Now, As tangents drawn from an external point are equal.

We have

AP = AS

[tangents from point A]

BP = BQ

[tangents from point B]

CR = CQ

[tangents from point C]

DR = DS

[tangents from point D]

Add the above equations

AP + BP + CR + DR = AS + BQ + CQ + DS

AB + CD = AS + DS + BQ + CQ

AB + CD = AD + BC

AB + AB = BC + BC [From 1]

AB = BC …[2]

From [1] and [2]

AB = BC = CD = AD

And we know,

A parallelogram with all sides equal is a rhombus

So, reason is also true, but not a correct reason for assertion.

Hence, B is correct option .

For Assertion :

In the given Figure,

As tangents drawn from an external point are equal.

We have

AP = AS

[tangents from point A]

BP = BQ

[tangents from point B]

CR = CQ

[tangents from point C]

DR = DS

[tangents from point D]

Add the above equations

AP + BP + CR + DR = AS + BQ + CQ + DS

AB + CD = AS + DS + BQ + CQ

AB + CD = AD + BC

So, assertion is not true

For Reason,

Consider two concentric circles with common center O and AB is a chord to outer circle and is tangent to inner circle P.

Now,

OP ⏊ AB

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

We know, that perpendicular from center to chord bisects the chord.

So, P bisects AB.

Reason is true

Hence, Assertion is false, But Reason is true.

In the given figure PT is a tangent to circle ∴ we have

∠OPT = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OPQ + ∠QPT = 90°

∠OPQ + 50° = 90°

∠OPQ = 40°

Now, In △POQ

OP = OQ

∠PQO = ∠QPO = 40°

[Angles opposite to equal sides are equal]

Now,

∠ PQO + ∠QPO + ∠ POQ = 180° [

By angle sum property of triangle]

40° + 40° + ∠POQ = 180°

∠POQ = 100°

Let us consider a circle with center O and OA and OB are two radii such that ∠AOB = 60° .

AP and BP are two intersecting tangents at point P at point A and B respectively on the circle .

To find : Angle between tangents, i.e. ∠APB

As AP and BP are tangents to given circle,

We have,

OA ⏊ AP and OB ⏊ BP

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠OAP = ∠OBP = 90°

In quadrilateral AOBP,

By angle sum property of quadrilateral, we have

∠OAP + ∠OBP + ∠APB + ∠AOB = 360°

90° + 90° + ∠APB + 130° = 360°

∠APB = 50°

In △AOP and △BOP

AP = BP

[Tangents drawn from an external point are equal]

OP = OP [Common]

OA = OB

[Radii of same circle]

△AOP ≅ △BOP

[By Side - Side - Side criterion]

∠APO = ∠BPO

[Corresponding parts of congruent triangles are congruent]

∠APB = ∠APO + ∠BPO

80 = ∠APO + ∠APO

2∠APO = 80

∠APO = 40°

In △AOP

∠APO + ∠AOP + ∠OAP = 180°

[By angle sum property]

40° + ∠AOP + 90° = 180°

[ ∠OAP = 90° as OA ⏊ AP because Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠AOP = 50°

Given : From an external point A, two tangents, AD and AE are drawn to a circle with center O. At a point F on the circle tangent is drawn which intersects AE and AD at B and C, respectively. And AE = 5 cm

To Find : Perimeter of △ABC

As we know that, Tangents drawn from an external point to a circle are equal.

So we have

BE = BF …[1]

[Tangents from point B]

CF = CD …[2]

[Tangents from point C]

Now Perimeter of Triangle abc

= AB + BC + AC

= AB + BF + CF + AC

= AB + BE + CD + AC …[From 1 and 2]

= AE + AD

Now,

AE = AD = 5 cm as tangents drawn from an external point to a circle are equal

So we have

Perimeter = AE + AD = 5 + 5 = 10 cm

As we know, Tangents drawn from an external point are equal.

CR = CQ [tangents from point C]

CQ = 3 cm [as CR = 3 cm]

Also,

BC = BQ + CQ

7 = BQ + 3 [BC = 7 cm]

BQ = 4 cm

Now,

BP = BQ [tangents from point B]

BP = 4 cm …[1]

AP = AS [tangents from point A]

AP = 5 cm [As AC = 5 cm] ….[2]

AB = AP + BP = 5 + 4 = 9 cm [From 1 and 2]

AB = x = 9 cm

Given : PA and PB are tangents to a circle with center O

To show : A, O, B and P are concyclic i.e. they lie on a circle i.e. AOBP is a cyclic quadrilateral.

Proof :

OB ⏊ PB and OA ⏊ AP

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

∠OBP = ∠OAP = 90°

∠OBP + ∠OAP = 90 + 90 = 180°

AOBP is a cyclic quadrilateral i.e. A, O, B and P are concyclic.

[ As we know if the sum of opposite angles in a quadrilateral is 180° then quadrilateral is cyclic ]

Hence Proved.

In the given Figure,

PA = PB

[Tangents drawn from an external points are equal]

∠PBA = ∠PAB

[Angles opposite to equal sides are equal]

∠PBA = ∠PAB = 65°

In △APB

∠PAB + ∠PBA + ∠APB = 180°

65° + 65° + ∠APB = 180°

∠APB = 50°

Also,

OB ⏊ AP

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OAP = 90°

∠OAB + ∠PAB = 90°

∠OAB + 65° = 90°

∠OAB = 25°

Given : A circle with center O , BC and BD are two tangents such that ∠CBD = 120°

To Proof : OB = 2BC

Proof :

In △BOC and △BOD

BC = BD

[Tangents drawn from an external point are equal]

OB = OB

[Common]

OC = OD

[Radii of same circle]

△BOC ≅ △BOD [By Side - Side - Side criterion]

∠OBC = ∠OBD

[Corresponding parts of congruent triangles are congruent]

∠OBC + ∠OBD = ∠CBD

∠OBC + ∠OBC = 120°

2 ∠OBC = 120°

∠OBC = 60°

In △OBC

⇒OB = 2BC

Hence Proved !

(i) secant

(ii) two

(iii) point of contact

(iv) infinitely many

Let us consider a circle with center O.

TP and TQ are two tangents from point T to the circle.

To Proof : PT = QT

Proof :

OP ⏊ PT and OQ ⏊ QT

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OPT = ∠OQT = 90°

In △TOP and △QOT

∠OPT = ∠OQT

[Both 90°]

OP = OQ

[Common]

OT = OT

[Radii of same circle]

△TOP ≅ △QOT

[By Right Angle - Hypotenuse - Side criterion]

PT = QT

[Corresponding parts of congruent triangles are congruent]

Hence Proved.

Let AB be the diameter of a circle with center O.

CD and EF are two tangents at ends A and B respectively.

To Prove : CD || EF

Proof :

OA ⏊ CD and OB ⏊ EF

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OAD = ∠OBE = 90°

∠OAD + ∠OBE = 90° + 90° = 180°

Considering AB as a transversal

⇒ CD || EF

[Two sides are parallel, if any pair of the interior angles on the same sides of transversal is supplementary]

We know, that tangents drawn from an external point are equal.

AD = AF

[tangents from point A] [1]

BD = BE

[tangents from point B] [2]

CF = CE

[tangents from point C] [3]

Now,

AB = AC [Given] …[4]

Substracting [1] From [4]

AB - AD = AC - AF

BD = CF

BE = CE [From 2 and 3]

Hence Proved.

Let PT and PQ are two tangents from external point P to a circle with center O

To Prove : PT and PQ subtends equal angles at center i.e. ∠POT = ∠QOT

In △OPT and △OQT

OP = OQ [radii of same circle]

OT = OT [common]

PT = PQ [Tangents drawn from an external point are equal]

△OPT ≅ △OQT [By Side - Side - Side Criterion]

∠POT = ∠QOT [Corresponding parts of congruent triangles are congruent]

Hence, Proved.

Let us consider a circle with center O and BC be a chord, and AB and AC are tangents drawn at end of a chord

To Prove : AB and AC make equal angles with chord, i.e. ∠ABC = ∠ACB

Proof :

In △ABC

AB = PC

[Tangents drawn from an external point to a circle are equal]

∠ACB = ∠ABC

[Angles opposite to equal sides are equal]

Hence Proved.

Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.

To Proof : ABCD is a rhombus.

As ABCD is a parallelogram

AB = CD and BC = AD …[1]

[opposite sides of a parallelogram are equal]

Now, As tangents drawn from an external point are equal.

We have

AP = AS

[tangents from point A]

BP = BQ

[tangents from point B]

CR = CQ

[tangents from point C]

DR = DS

[tangents from point D]

Add the above equations

AP + BP + CR + DR = AS + BQ + CQ + DS

AB + CD = AS + DS + BQ + CQ

AB + CD = AD + BC

AB + AB = BC + BC [From 1]

AB = BC …[2]

From [1] and [2]

AB = BC = CD = AD

And we know,

A parallelogram with all sides equal is a rhombus

So, ABCD is a rhombus.

Hence Proved.

Given : Two concentric circles (say C₁ and C₂) with common center as O and radius r₁ = 5 cm and r₂ = 3 cm respectively.

To Find : Length of the chord of the larger circle which touches the circle C₂. i.e. Length of AB.

As AB is tangent to circle C₂ and we know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"

So, we have,

OP ⏊ AB

∴ OPB is a right - angled triangle at P,

By Pythagoras Theorem in △OPB

[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]

We have,

(OP)² + (PB)² = (OB)²

r₂² + (PB)² = r₁²

(3)² + (PB)²= (5)²

9 + (PB)² = 25

(PB)² = 16

PB = 4 cm

Now, AP = PB ,

[as perpendicular from center to chord bisects the chord and OP ⏊ AB]

So,

AB = AP + PB = PB + PB = 2PB = 2(4) = 8 cm

Let us consider a quadrilateral ABCD, And a circle is circumscribed by ABCD

Also, Sides AB, BC, CD and DA touch circle at P, Q, R and S respectively.

To Proof : Sum of opposite sides are equal, i.e. AB + CD = AD + BC

Proof :

In the Figure,

As tangents drawn from an external point are equal.

We have

AP = AS

[tangents from point A]

BP = BQ

[tangents from point B]

CR = CQ

[tangents from point C]

DR = DS

[tangents from point D]

Add the above equations

AP + BP + CR + DR = AS + BQ + CQ + DS

AB + CD = AS + DS + BQ + CQ

AB + CD = AD + BC

Hence Proved.

Consider a quadrilateral, ABCD circumscribing a circle with center O and AB, BC, CD and AD touch the circles at point P, Q, R and S respectively.

Joined OP, OQ, OR and OS and renamed the angles (as in diagram)

To Prove : Opposite sides subtends supplementary angles at center i.e.

∠AOB + ∠COD = 180° and ∠BOC + ∠AOD = 180°

Proof :

In △AOP and △AOS

AP = AS

[Tangents drawn from an external point are equal]

AO = AO

[Common]

OP = OS

[Radii of same circle]

△AOP ≅ △AOS

[By Side - Side - Side Criterion]

∠AOP = ∠AOS

[Corresponding parts of congruent triangles are congruent]

∠1 = ∠2 …[1]

Similarly, We can Prove

∠3 = ∠4 ….[2]

∠5 = ∠6 ….[3]

∠7 = ∠8 ….[4]

Now,

As the angle around a point is 360°

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°

∠2 + ∠2 + ∠3 + ∠3 + ∠6 + ∠6 + ∠7 + ∠7 = 360° [From 1, 2, 3 and 4]

2(∠2 + ∠3 + ∠6 + ∠7) = 360°

∠AOB + ∠COD = 180°

[As, ∠2 + ∠3 = ∠AOB and ∠5 + ∠6 = ∠COD] [5]

Also,

∠AOB + ∠BOC + ∠COD + ∠AOD = 360°

[Angle around a point is 360°]

∠AOB + ∠COD + ∠BOC + ∠AOD = 360°

180° + ∠BOC + ∠AOD = 360° [From 5]

∠BOC + ∠AOD = 180°

Hence Proved

Let us consider a circle with center O and PA and PB are two tangents to the circle from an external point P

To Prove : Angle between two tangents is supplementary to the angle subtended by the line segments joining the points of contact at center, i.e. ∠APB + ∠AOB = 180°

Proof :

As AP and BP are tangents to given circle,

We have,

OA ⏊ AP and OB ⏊ BP

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠OAP = ∠OBP = 90°

In quadrilateral AOBP, By angle sum property of quadrilateral, we have

∠OAP + ∠OBP + ∠AOB + ∠APB = 360°

90° + 90° + ∠AOB + ∠APB = 360°

∠AOB + ∠APB = 180°

Hence Proved

Given : A circle with center O and radius 3 cm and PQ is a chord of length 4.8 cm. The tangents at P and Q intersect at point T

To Find : Length of PT

Construction : Join OQ

Now in △OPT and △OQT

OP = OQ

[radii of same circle]

PT = PQ

[tangents drawn from an external point to a circle are equal]

OT = OT

[Common]

△OPT ≅ △OQT

[By Side - Side - Side Criterion]

∠POT = ∠OQT

[Corresponding parts of congruent triangles are congruent]

or ∠POR = ∠OQR

Now in △OPR and △OQR

OP = OQ

[radii of same circle]

OR = OR [Common]

∠POR = ∠OQR [Proved Above]

△OPR ≅ △OQT

[By Side - Angle - Side Criterion]

∠ORP = ∠ORQ

[Corresponding parts of congruent triangles are congruent]

Now,

∠ORP + ∠ORQ = 180°

[Linear Pair]

∠ORP + ∠ORP = 180°

∠ORP = 90°

⇒ OR ⏊ PQ

⇒ RT ⏊ PQ

As OR ⏊ PQ and Perpendicular from center to a chord bisects the chord we have

PR = QR = PQ/2 = 16/2 = 8 cm

∴ In right - angled △OPR,

By Pythagoras Theorem

[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]

(OP)² = (OR)² + (PR)²

(10)² = (OR)² + (8)²

100 = (OR)² + 64

(OR)²= 36

OR = 6 cm

Now,

In right angled △TPR, By Pythagoras Theorem

(PT)² = (PR)² + (TR)² [1]

Also, OP ⏊ OT

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

In right angled △OPT, By Pythagoras Theorem

(PT)² + (OP)² = (OT)²

(PR)² + (TR)² + (OP)²= (TR + OR)² [From 1]

(8)² + (TR)² + (10)² = (TR + 6)²

64 + (TR)² + 100 = (TR)² + 2(6)TR + (6)²

164 = 12TR + 36

12TR = 128

TR = 10.7 cm [Appx]

Using this in [1]

PT² = (8)² + (10.7)²

PT² = 64 + 114.49

PT² = 178.49

PT = 13.67 cm [Appx]