Tangents And Secants To A Circle

i. One

Property- a tangent to a circle touches it at only one point called the common point.

ii. Secant

Secant- a line that touches the circle at two different points

ii. Two

A circle can have two tangents that are parallel, these two tangents touch the circle on the opposite sides, and distance between the point of contacts is the diameter.

iv. Point of contact

The point where tangent touches the circle is called point of contact.

v. Infinite

We can draw infinite tangents to a circle, as a circle can be assumed to be curve having infinite points and one tangent can be drawn from each point on the circle.

We know that tangent to a circle makes a right angle with radius.

∠OPQ = 90°

Applying Pythagoras

PQ² = OP² + OQ²

PQ² = 5² + 12²

PQ² = 25 + 144

PQ² = 169

PQ = 13cm

Step1: Draw circle of random radius with BC as diameter.

Step2: Draw line AD perpendicular to BC to touch the circle at D

Such that ∠D = 90°

Step3: Draw line through D parallel to BC that will form a tangent

Step4: Take a random point on line AD as F and draw a line through F parallel to BC to intersect circle in two points that forms a secant.

We know that tangent to a circle makes a right angle with radius.

∠OPQ = 90°

Applying Pythagoras

PQ² = OP² + OQ²

PQ² = 9² + 15²

PQ² = 81 + 225

PQ² = 306

PQ = cm

Length of tangent = cm

To prove: DE ∥ FG

Proof:

We know that tangent to a circle makes a right angle with the radius.

Let DE and FG be tangent at B and C respectively.

BC forms the diameter.

∴ ∠OBE = ∠OBD = ∠OCG = ∠OCF = 90°

Also, ∠OBD = ∠OCG and ∠OBE = ∠OCF as alternate angles

∴ DE and FG make 90° to same line BC which is the diameter.

Thus DE ∥ FG

Property of tangent- tangent to a circle makes right angle with radius at point of contact

Let O be center, OP be radius.

Applying Pythagoras

PQ = 24, OQ = 25

OQ² = OP² + PQ²

25² = OP² + 24²

625 = OP² + 576

OP² = 49

OP = 7 cm

Radius of Circle = 7 cm

.

We know that tangent to a circle makes right angle with radius.

∴ ∠A = ∠B = 90° and ∠P = 80°

Sum of all angles of quadrilateral = 180°

∴ ∠A + ∠B + ∠P + ∠O = 360°

∴ 90° + 90° + 80° + ∠O = 360°

∠O = 100°

∠ AOP = ∠ BOP

∴ ∠POA = 50°

Construct Line OC = radius

OP = OQ = OC = radius

OC∥AP and AC∥OP and

Thus AP = OP = radius

As AP = OP and ∠P = 90°

ΔOAP is isosceles triangle

∴ ∠ PAO = ∠POA = 45°

Also ∠ PAC = 90°

∠OAC = 45° ---1

Also AB is perpendicular to OC

∠OCA = 90°

In ΔAOC

∠OCA + ∠OAC + ∠COA = 180°

45 + 90 + ∠COA = 180°

∠COA = 45°

Similarly

∠BOC = 45°

∴ ∠AOB = 90°

AB = AF = 5cm radius of larger circle

AD = AC = 3cm radius of smaller circle

Pythagoras theorem

AF² = AD² + DF²

5² = 3² + DF²

25² = 9² + DF²

DF = 4 units

Length of chord = 2 × DF = 2 × 4 = 8cm

FGHI is a parallelogram

∴ HI = FG and FI = GH----1

Also

IB = IE tangents to circle from I

And similarly

HE = HD, CG = GD and CF = BF

Adding all the equations

IE + HE + GC + CF = BF + BI + GD + DH

IH + GF = IF + GH

∴ 2HI = 2HG

HI = HG---2

∴ GF = GH = HI = IF

From 1 and 2

Thus FGHI is a rhombus

Construction : Draw radius OB = 3cm

Proof: AC, BC and AB are tangents

BF = BD = 9cm—tangents from B

AF = AB—tangent from A

∴ CD = CB tangents from C

OD = OB = 3cm radius

OBCD forms a square of side 3cm

OD = OB = BC = CD = 3cm

∴ ∠BCD = 90°

BD = 9cm and DC = 3cm

OB = 3cm radius of circle

Let AF = AB = x

Applying Pythagoras

AB = BC + AC

(9 + x)² = (12)² + (3 + x)²

81 + 18x + x² = 144 + 9 + 6x + x²

12x = 72

X = 6cm

∴ AB = 9 + 6 = 15cm

AC = 6 + 3 = 9cm

Step1: Draw circle of radius 6cm with center A, mark point C at 10 cm from center

Step 2: find perpendicular bisector of AC

Step3: Take this point as center and draw a circle through A and C

Step4:Mark the point where this circle intersects our circle and draw tangents through C

Length of tangents = 8cm

AE is perpendicular to CE (tangent and radius relation)

In ΔACE

AC becomes hypotenuse

AC² = CE² + AE²

10² = CE² + 6²

CE² = 100-36

CE² = 64

CE = 8cm

Step1:Draw circles of radius 4 and 6 cm

Step 3: Draw tangent to inner circle from C

AD is perpendicular to DC- tangent and radius

In Δ ADC

AC is radius

AC² = AD² + DC²

6² = 4² + DC²

36 = 16 + DC²

DC² = 20

DC = 2 cm

Step1: Draw a circle with bangle, its center is not known

Step2: to find the center draw 2 random chords C and FE

Step 3:Find the perpendicular bisectors of these chords, the points where they intersect is the center of circle

Step 4:

Draw a line from center to a point outside it.

Step 5: draw its perpendicular bisector

Step 6: Cut arcs over the circle with this point as center and point outside it as radius

Step 7: Draw tangent to the circle where the arcs cut the circle

ΔABC is right angled triangle

∠ ABC = 90°

Drawn with AB as diameter that intersects AC at P, PQ is the tangent to the circle which intersects BC at Q.

Join BP.

PQ and BQ are tangents from an external point Q.

∴ PQ = BQ ---1 tangent from external point

⇒ ∠PBQ = ∠BPQ = 45° (isosceles triangle)

Given that, AB is the diameter of the circle.

∴ ∠APB = 90° (Angle subtended by diameter)

∠APB + ∠BPC = 180° (Linear pair)

∴ ∠BPC = 180° – ∠APB = 180° – 90° = 90°

Consider ΔBPC,

∠BPC + ∠PBC + ∠PCB = 180° (Angle sum property of a triangle)

∴ ∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90° ---2

∠BPC = 90°

∴ ∠BPQ + ∠CPQ = 90° ...3

From equations 2 and 3, we get

∠PBC + ∠PCB = ∠BPQ + ∠CPQ

⇒ ∠PCQ = ∠CPQ (Since, ∠BPQ = ∠PBQ)

Consider ΔPQC,

∠PCQ = ∠CPQ

∴ PQ = QC ---4

From equations 1 and 4, we get

BQ = QC

Therefore, tangent at P bisects the side BC.

Let O be center of circle and R be a point outside it.

Draw tangent to circle at A and B through R

i. Let Major segment A₁ minor segment be A₂

∠A = 90°

area of sector ACD = 78.5cm²

Pythagoras

AC² = CD² + AD²

AC² = 10² + 10²

AC² = 200

AC = 10

Height of triangle =

Cos 45° =

AE =

Area of minor segment = Area of sector ACD – Area ΔACD

= 78.5 –10

= 78.5-50

= 28.5 cm²

ii. Major segment = Area of circle – minor segment

= π × r² –minor segment

= π × 10² –28.5

= 314-28.5

= 285.5cm²

Let Major segment A₁ minor segment be A₂

area of sector ACD = 150.72 sq.cm

Sin 30° =

AE = 6cm

Cos 30° =

DE = 1.732 × 6

DE = 10.392cm

CD = 2 × 10.392 = 20.784cm

Area of minor segment = Area of sector ACD – Area ΔACD

= 150.72 –20.784

= 1550.72-62.352

= 88.44 cm²

Radius = 25cm

Angle = 115°

Area of sector ACD = 627.22cm2

For 2 such wipers: 2 × 627.22

Area = 1254.45 cm2

Consider midpoint side AB, it forms Smaller square of size 5cm

For this smaller square

Area of shaded region

= Area of 1^st quadrant + area of 2^nd quadrant –area of square

Area of both quadrants is same

∴ Area of shaded region

= 2 × Area of quadrant - area of square

= 2 × – r × r

= 2 × – 5 × 5

= 2 × – 5 × 5

= 14.25 cm²

Area of total shaded region = 4 × 14.25

= 57 cm²

Area of shaded region = Area of square – area of 2 semicircles

Area of square = 7 × 7 = 49 sq.cm

Area of semicircles =

= π × r²

= π × 3.5²

= 38.5 sq.cm

Area of shaded region = 49-38.5 = 10.5 cm²

Area of Shaded region = Area of sector OABC-Area of ΔDOB

OB = OA = 3.5cm

area of sector OABC = 9.625 cm²

Area of ΔDOB =

= 3.5 cm²

Area of Shaded region = 9.625-3.5

= 6.125 cm²

Area of shaded region = Area(Sector OAB)-Area(Sector OCD)

= 102.67 cm²

Area of shaded region

= Area of 1^st quadrant + area of 2^nd quadrant –area of square

Area of both quadrants is same

∴ Area of shaded region

= 2 × Area of quadrant - area of square

= 2 × – r × r

= 2 × – 10 × 10

= 2 × – 10 × 10

= 57.07 cm²