Circle

(i) The centre of the circle lies in interior of the circle.

We know that the centre is a fixed point in a circle i.e. interior of the circle.

(ii) A point, where distance from the centre of a circle is greater than its radius lies in exterior of the circle.

If the point’s distance from centre of circle is equals to its radius, then it lies on the circle.

If the distance is greater than the radius, then it lies in exterior of the circle.

(iii) The longest chord of a circle is a diameter of the circle.

A chord that passes through the centre of the circle is the longest chord and therefore it is the diameter.

(iv) An arc is a semicircle when its ends are the ends of the diameter.

Let AB be the diameter and AXB be the arc.

Here, we can see that AXB is a semicircle.

Thus, an arc is a semicircle when its ends are the ends of a diameter.

(v) A circle divides the plane, on which it lies, in three parts.

We know that a circle divides the plane on which it lies into three parts i.e.

1. Interior

2. Circle

3. Exterior

(i) True

We know that the constant distance is radius of a circle.

The line segment joining the centre to any point on the circle has constant distance.

∴ Line segment joining the centre to any point on the circle is a radius of the circle.

(ii) False

In a circle, infinitely many diameters can be drawn.

A diameter is the longest chord.

∴ a circle has infinitely many chords.

(iii) False

When a circle is divided into three equal parts, each part will be less than a semicircle.

(iv) True

We know that diameter is twice the radius i.e. d = 2r.

∴ A chord of a circle, which is twice as long as its radius is a diameter of the circle.

(v) True

A circle is drawn on a plane which divides it into three parts.

∴ A circle is a plane figure.

(vi) False

The collection of those points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle.

(vii) False

We know that if any chord passes through the centre of the circle then it is called the diameter of the circle.

(i) False

We know that longer chords subtend at greater angles and smaller chords at smaller angles.

Here CD (longer chord) is subtending at 50° (smaller angle) while AB (smaller chord) is subtending at 70° (greater angle).

So, it is false.

(ii) False

We know that longer the chord, smaller the distance from centre.

Here, 10 cm chord is at a distance of 8 cm from the center, hence it is not possible that 8 cm chord is at a distance of 5 cm from the center.

(iii) True

We know that if two chords are at equal distance from the centre, they are equal.

Here, chords AB and CD are at equal distance of 4 cm from the centre.

∴ AB = CD

Hence, it is true.

(iv) True

We know that equal chords of congruent circles subtend equal angles at the corresponding centres.

If we take AB as chord, and as radius of both circle are equal

⇒ ∠AOB = ∠AO’B

So, it is true.

(v) False

A circle passing through two collinear points cannot pass through the third point.

Hence, it is false.

(vi) True

Given radius = 4 cm

We know that diameter is twice the radius.

When a circle is drawn through two points A and B, diameter will be AB = 8 cm.

Taking a compass, from the centre O and OA or OB as radius, we join A to B and B to A.

So, the circle can be drawn through A and B.

∴ It is true.

We know that Length of chord,

Where r = radius of circle, d = distance of chord from centre

Given radius = 13 cm and chord = 10 cm

Now,

Squaring on both sides,

⇒ 100 = 4 (169 – d²)

⇒ 100/ 4 = 169 – d²

⇒ 25 = 169 – d²

⇒ d² = 169 – 25

⇒ d² = 144

∴ Distance of chord, d = 12 cm

Given chords AB = 6 cm and CD = 12 cm,

Draw OE⊥ AB intersecting CD at F,

As, AB||CD

OF⊥ CD [Corresponding angles]

Distance between AB and CD, EF= 3 cm

Now, we know that perpendicular from center to chord bisects the chord.

Then CF = FD = 1/2 CD = 6 cm

⇒ AE = EB = 1/2 AB = 3 cm

Let OF = y cm, OE = 3 + y cm and OD = OB = x cm = Radius.

Consider ΔOFD,

By Pythagoras Theorem,

⇒ OD² = OF² + FD²

⇒x² = y² + 6²

⇒x² = y² + 36 … (1)

Consider ΔOEB,

By Pythagoras Theorem,

⇒ OB² = OE² + EB²

⇒x² = (3 + y)² + 3²

⇒x² = 9 + y² + 6y + 9

⇒x² = y² + 6y + 18 … (2)

From (1) and (2),

⇒ y² + 36 = y² + 6y + 18

⇒ 36 = 6y + 18

⇒ 6y = 36 – 18

⇒ 6y = 18

⇒ y = 18/ 6

∴ y = 3

Substituting y value in (1),

⇒x² = y² + 36

⇒x² = 3² + 36

⇒x² = 9 + 36

⇒x² = 45

[45 = 3 × 3 × 5]

∴ Radius = x = 3√5 cm

Given chord AB = chord CD intersect at E.

Construction:

Join AD and BC.

Draw OF ⊥ AB and OG ⊥ CD.

Join OE.

We know that when a perpendicular is drawn from centre to chord, it bisects them.

⇒ AF = FB = and CG = GD =

Given AB = CD,

⇒ AF = FB = CG = GD … (1)

Consider ΔOFE and ΔOGE,

We know that equal chords of a circle are equidistant from each other.

⇒ OF = OG

⇒ OE = OE [Common side]

⇒ ∠OFE = ∠OGE = 90° [Construction]

By RHS congruence rule,

⇒ ΔOFE ≅ ΔOGE

By CPCT,

⇒ FE = GE … (2)

So,

⇒ AE = AF + FE, CE = CG + GE

And EB = FB – FE, ED = GD – GE

From (1) and (2),

⇒ AE = CE … (3)

And EB = ED … (4)

Consider ΔAED and ΔCEB,

⇒ AE = CE [From (3)]

⇒ EB = ED [From (4)]

⇒ ∠AED = ∠CEB [Vertically opposite angles are equal]

By SAS congruence rule,

⇒ ΔAED ≅ ΔCEB

By CPCT,

⇒ AD = CB

∴ Arc AD = Arc CB

Hence proved

Given AB and CD are equal chords of a circle with centre O.

Also OM ⊥ AB and ON ⊥ CD.

We have to prove that ∠OMN = ∠ONM.

Proof:

We know that equal chords of a circle are equidistant from the centre.

⇒ OM = ON … (1)

In ΔOMN,

From (1),

⇒ OM = ON

We know that angles opposite to equal sides of a triangle are equal.

∴ ∠OMN = ∠ONM

Hence proved.

Given O and O’ are the centres of the given circles.

Also AB || OO’

We have to prove that AB = 2OO’.

Construction:

Draw perpendicular CP from point C on OO’.

Proof:

We know that perpendicular from the centre of a circle to a chord bisects the chord.

⇒ BE = EC and CD = DA

Since AB || OO’ and O’E || PC || OD, [since all are perpendiculars on line AB]

⇒ EC = O’P and CD = PO

⇒ OO’ = OP + PO’

= CD + EC

= 1/2 BC + 1/2 AC

= 1/2 (BC + AC)

= 1/2 AB

∴ 2OO’ = AB

Hence proved.

Given chords AB = 10 cm, CD = 24 cm

Distance between AB and CD = 17 cm

From the figure,

OE ⊥ AB and OF ⊥CD.

Now, we know perpendicular from the center to the chord bisects the chord.

⇒ AE = EB = 1/2 AB = 5 cm

⇒ CF = FD = 1/2 CD = 12 cm

Let OF = y cm, OE = 17 - y cm and OB = OD = x cm = Radius.

Consider ΔOEB,

By Pythagoras Theorem,

⇒ OB² = OE² + EB²

⇒x² = (17 – y)² + 5²

⇒x² = 289 + y² – 34y + 25

⇒x² = y² – 34y + 314 … (1)

Consider ΔOFD,

By Pythagoras Theorem,

⇒ OD² = OF² + FD²

⇒x² = y² + 12²

⇒x² = y² + 144 … (2)

From (1) and (2),

⇒ y² – 34y + 314 = y² + 144

⇒ -34y + 314 = 144

⇒ 34y = 314 – 144

⇒ 34y = 170

⇒ y = 170/ 34

∴ y = 5

Substituting y value in (2),

⇒x² = 5² + 144

⇒x² = 25 + 144

⇒x² = 169

∴ Radius = x = 13 cm

Given radius of circle = 10 cm

Chords AB = 12 cm and CD = 16 cm

(a)

Draw OE, OF⊥ AB, CD.

We know that when a perpendicular is drawn from centre to chord, it bisects them.

Then CF = FD = 1/2 CD = 8 cm

⇒ AE = EB = 1/2 AB = 6 cm

Consider ΔOFD,

By Pythagoras Theorem,

⇒ OD² = OF² + FD²

⇒ 10² = x² + 8²

⇒100² = x² + 64

⇒x² = 100 – 64

⇒x² = 36

∴ OF = x = 6 cm

Consider ΔOEB,

By Pythagoras Theorem,

⇒ OB² = OE² + EB²

⇒10² = y² + 6²

⇒100 = y² + 36

⇒y² = 100 – 36

⇒ y² = 64

∴ OE = y = 8 cm

∴ Distance between chords AB and CD = EF = OE – OF = y – x = 8 – 6 = 2 cm

(b)

Draw OE, OF ⊥ AB, CD.

We know that when a perpendicular is drawn from centre to chord, it bisects them.

Then CF = FD = 1/2 CD = 8 cm

⇒ AE = EB = 1/2 AB = 6 cm

Consider ΔOFD,

By Pythagoras Theorem,

⇒ OD² = OF² + FD²

⇒ 10² = x² + 8²

⇒ 100² = x² + 64

⇒x² = 100 – 64

⇒x² = 36

∴ OF = x = 6 cm

Consider ΔOEB,

By Pythagoras Theorem,

⇒ OB² = OE² + EB²

⇒ 10² = y² + 6²

⇒ 100 = y² + 36

⇒ y² = 100 – 36

⇒ y² = 64

∴ OE = y = 8 cm

∴ Distance between chords AB and CD is

⇒ EF = OE + OF

= 8 + 6

= 14 cm

Given AB = CD.

Construction:

Join AD and BC.

Consider ΔABC and ΔBCD,

⇒ AB = CD [Given]

⇒ ∠ABC = ∠BCD [Vertically opposite angles are equal]

⇒ BC = BC [Common side]

By SAS congruence rule,

⇒ ΔABC ≅ ΔBCD

By CPCT,

⇒ AC = BD

Hence proved.

Let AB and CD be two equal chords of a circle which are intersecting at a point E.

Construction:

Draw OF and OG perpendiculars on the chords

Join OE.

Consider ΔOFE and ΔOGE,

⇒ OF = OG [Equal chords]

⇒∠OFE = ∠OGE [Each 90°]

⇒ OE = OE [Common]

By RHS congruence rule,

⇒ ΔOFE ≅ ΔOGE

By CPCT,

⇒ FE = GE … (1)

Given AB = CD … (2)

⇒ 1/2 AB = 1/2 CD

⇒ AG = CF … (3)

Adding equations (1) and (3),

⇒ AG + GE = CF + FE

⇒ AE = CE … (4)

Subtracting equation (4) from (2),

⇒ AB – AE = CD - CE

⇒ BE = DE … (5)

From (4) and (5),

We can see that two parts of one chord are equal to two parts of another chord.

Hence proved

Let AB and CD be the two parallel chords of a circle such that M and N are the mid-points of AB and CD respectively.

Since the perpendicular bisector of the chord passes through the centre,

⇒ ON ⊥ CD and OM ⊥ AB

Since AB || CD, NOM is a straight line.

Hence the line joining the midpoints of two parallel chords of a circle passes through the centre of the circle.

(i) False

We know that if two points lie on the same segment, then they are equal.

Hence, it is false.

(ii) False

⇒ ∠C = 90°

∴ By Pythagoras Theorem,

⇒ AB² = AC² + BC²

(iii) True

Join AD, DE, EB and DB.

∠ADB = 90°

And ∠BDE = 120° - 90° = 30°

We know that angles that lie on the same segment are equal.

∴ ∠BDE = ∠EAB = 30°

(iv) True

Join AC, CE, CD, DE, AE and AD.

ACDE is minor segment of the circle.

We know that angles in the same segment are equal.

∴ ∠CAD = ∠CED

We know that angle subtended by an arc at the centre is double the angle subtended at any point on the remaining part of the circle.

⇒∠AOC = 2∠ABC

Given ∠ABC = 45°

⇒∠AOC = 2(45°)

⇒∠AOC = 90°

Since ∠AOC = 90°, OA ⊥ OC.

Hence proved.

Given in ΔABC, O is the circum circle and D is the mid-point of the base BC.

We have to prove that ∠BOD = ∠A.

Construction:

Join OB and OC

Proof:

Consider ΔOBD and ΔOCD,

⇒ OB = OC [Radii of same circle]

⇒ BD = DC [D is mid-point of BC]

⇒ OD = OD [Common side]

By SSS congruence rule,

ΔOBD ≅ ΔOCD

By CPCT,

⇒ ∠BOD = ∠COD

∴ ∠BOD = 1/2 ∠BOC … (1)

Arc BC subtends ∠BOC at the centre and ∠BAC at point A in the remaining part of the circle.

∴ ∠BAC = 1/2 ∠BOC … (2)

From (1) and (2),

⇒∠BOD = ∠BAC

∴ ∠BOD = ∠A

Hence proved

Given ACB and ADB are two right angled triangles having common hypotenuse AB.

We have to prove that ∠BAC = ∠BDC.

Construction: Join CD.

Proof:

⇒∠C + ∠D = 90° + 90° = 180°

We know that if opposite angles of a quadrilateral are supplementary, then it is a cyclic quadrilateral.

∴ ADBC is a cyclic quadrilateral.

We know that if two angles are of the same arc, then they are equal.

Here, ∠BAC and ∠BDC are made by the same arc BC.

∴ ∠BAC = ∠BDC

Hence proved

Given AB subtends at an angle 90° and AC subtends at 150°.

We know that sum of all angles at a point = 360°

⇒∠AOC + ∠AOB + ∠COB = 360°

⇒ 150° + 90° + ∠COB = 360°

⇒ 240° + ∠COB = 360°

⇒∠COB = 360° - 240°

∴ ∠COB = 120°

We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point.

⇒∠COB = 2 ∠CAB

⇒∠CAB = 1/2 ∠COB

= 1/2 (120°)

= 60°

∴ ∠BAC = 60°

Given O is the circum centre of ΔABC.

Construction:

Join OC

We have to prove that ∠OBC + ∠BAC = 90°.

Proof:

Consider ΔOBC,

⇒ OB = OC [radii of same circle]

We know that angles opposite to equal sides are equal

⇒ ∠OBC = ∠OCB … (1)

We know that angle at the center is twice the angle at the circumference.

⇒∠BOC = 2∠BAC … (2)

We know that sum of angles of a triangle is 180°.

⇒∠OBC + ∠OCB + ∠BOC = 180°

⇒ 2∠OBC + ∠BOC = 180° [From (1)]

⇒ 2∠OBC + 2∠BAC = 180°

⇒∠OBC + ∠BAC = 180°/ 2

∴ ∠OBC + ∠BAC = 90°

Hence proved

Given a chord of a circle is equal to its radius.

In ΔOAB,

⇒ AB = OA = OB.

∴ ΔOAB is an equilateral triangle.

Each angle of an equilateral triangle is 60°.

∴ ∠AOB = 60°

We know that angle subtended at the centre of a circle by an arc is double the angle subtended by it on any point on the remaining part of the circle.

⇒∠ACB = 1/2 ∠AOB = 1/2 (60°) = 30°

∴ Angle subtended by this chord at a point on the major arc is 30°.

Given ∠ADC = 130° and chord BC = chord BE.

Consider the points A, B, C and D which form a cyclic quadrilateral.

We know that in a cyclic quadrilateral the opposite angles are supplementary.

In cyclic quadrilateral ADCB,

⇒∠ADC + ∠OBC = 180°

⇒ 130° + ∠OBC = 180°

⇒∠OBC = 180° - 130° = 50°

Consider ΔBOC and ΔBOE,

⇒ BC = BE [given]

⇒ OC = OE [radii of same circle]

⇒ OB = OB [common side]

By SSS congruence rule,

⇒ ΔBOC ≅ ΔBOE

By CPCT,

⇒∠OBC = ∠OBE = 50°

⇒∠CBE = ∠CBO + ∠EBO

= 50° + 50°

∴ ∠CBE = 100°

Given ∠ACB = 40°

We know that a segment subtends an angle to the circle is half the angle that subtends to the circle.

i.e. ∠AOB = 2 ∠ACB

⇒∠AOB = 2 (40°)

⇒∠AOB = 80° … (1)

In ΔAOB,

⇒ OA = OB [Radii of same circle]

We know that angles opposite to equal sides are equal.

∴ ∠OBA = ∠OAB

We know that sum of all angles in a triangle is 180°.

⇒∠AOB + ∠OBA + ∠OAB = 180°

⇒ 80° + ∠OAB + ∠OAB = 180°

⇒ 2∠OAB = 180° - 80°

⇒ 2∠OAB = 100°

⇒∠OAB = 100°/ 2

∴ ∠OAB = 50°

ACDE is a cyclic quadrilateral because A, C, D and E are four points on circle.

Construction: Join AE

We know that in a cyclic quadrilateral the opposite angles are supplementary.

⇒∠ACD + ∠AED = 180° … (1)

Given AOB is a diameter.

We know that diameter subtends a right angle to the circle.

⇒∠AEB = 90°

On adding equations (1) and (2),

⇒∠ACD + ∠AED + ∠AEB = 180° + 90° = 270°

∴ ∠ACD + ∠BED = 270°

We know that the pair of opposite angles of a cyclic quadrilateral is supplementary i.e. sum is 180°.

Let the opposite angle of the cyclic quadrilateral be x°.

(i) Given angle = 70°

⇒ 70° + x° = 180°

⇒x° = 180° - 70°

∴ x° = 110°

(ii)Given angle = 135°

⇒ 135° + x° = 180°

⇒x° = 180° - 135°

∴ x° = 45°

(iii)Given angle = 1121/2°

⇒ 1121/2° + x° = 180°

⇒x° = 180° - 1121/2°

∴ x° = 671/2°

(iv)Given angle = right angle

⇒ 54° + x° = 180°

⇒x° = 180° - 54°

∴ x° = 126°

(v) Given angle = 165°

⇒ 165° + x° = 180°

⇒x° = 180° - 165°

∴ x° = 15°

We know that the pair of opposite angles of a cyclic quadrilateral is supplementary i.e. sum is 180°.

Let the angle be x°.

(i) of other

⇒

⇒

⇒

⇒ x° = 140°

And

∴ Angles are 140° and 40°.

(ii) of other

⇒

⇒

⇒

⇒ x° = 48°

And

∴ Angles are 48° and 132°.

We know that the pair of opposite angles of a cyclic quadrilateral is supplementary i.e. sum is 180°.

Firstly, 2x° + 3x° = 180°

⇒ 5x° = 180°

⇒x° = 180°/ 5

∴ x° = 36°

Also, y° + 2y° = 180°

⇒ 3y° = 180°

⇒y° = 180°/ 3

∴ y° = 60°

Now,

∴ ∠A = y° = 60°

∴ ∠B = 3x° = 3 (36°) = 108°

∴ ∠C = 2y° = 2 (60°) = 120°

∴ ∠D = 2x° = 2 (36°) = 72°

We know that in a cyclic quadrilateral, the exterior angle formed by producing one side of a quadrilateral is equal to its interior opposite angle.

∴ a = 65° and c = 91°

We know that the pair of opposite angles of a cyclic quadrilateral is supplementary i.e. sum is 180°.

Firstly, a + d = 180°

⇒ 65° + d = 180°

⇒ d = 180° - 65°

∴ d = 115°

Similarly, b + c = 180°

⇒ b + 91° = 180°

⇒ b = 180° - 91°

∴ b = 89°

∴ a = 65°, b = 89°, c = 91° and d = 115°

Given ABCD is a cyclic quadrilateral and AD || BC.

We have to prove that ∠A = ∠D.

Proof:

Since AD || BC and CD is a traversal.

So, ∠BCD + ∠ADC = 180° … (1)

But ABCD is a cyclic quadrilateral.

⇒∠BCD + ∠BAD = 180° … (2)

From (1) and (2),

⇒∠ADC = ∠BAD

∴ ∠D = ∠A

Hence proved.

Given that ABCD is a cyclic quadrilateral and AB and DC are produced to meet at E.

We have to prove that ΔEBC and ΔEDA are similar.

Proof:

We know that in a cyclic quadrilateral, opposite angles are supplementary.

⇒∠ABC + ∠ADC = 180° … (1)

Also, ∠ABC + ∠EBC = 180° … (2) [linear pair]

From (1) and (2),

⇒∠ABC + ∠ADC = ∠ABC + ∠EBC

⇒∠ADC = ∠EBC … (3)

Similarly,

⇒∠BAD + ∠BCD = 180° … (4)

Also, ∠BCD + ∠BCE = 180° … (5) [linear pair]

From (4) and (5),

⇒∠BAD + ∠BCD = ∠BCD + ∠BCE

⇒∠BAD = ∠BCE … (6)

And ∠BEC = ∠AED … (7)

From (3), (6) and (7),

∴ ΔEBC and ΔEDA are similar.

Hence proved

Given that ABCD is a cyclic quadrilateral.

We have to prove that EFGH is also a cyclic quadrilateral.

Proof:

Let

⇒ 1/2∠A = x; 1/2∠B = w; 1/2∠C = z and 1/2∠D = y

We know that opposite angles of a cyclic quadrilateral are supplementary.

⇒∠A + ∠C = 180° and ∠B + ∠D = 180°

⇒ 1/2 (A + C) = 90° and 1/2 (B + D) = 180°

⇒∠x + ∠z = 90° and ∠y + ∠w = 90° … (1)

In ΔAFD and ΔBHC,

⇒∠x + ∠y + ∠AFD = 180°

⇒∠AFD = 180° - (∠x + ∠y) … (2)

And ∠z + ∠w + ∠BHC = 180°

⇒∠BHC = 180° - (∠z + ∠w) … (3)

Adding (2) and (3),

⇒∠AGD + ∠BHC = 360° - (∠x + ∠y + ∠z + ∠w)

From (1),

⇒∠AGD + ∠BHC = 360° - 180°

∴ ∠AGD + ∠BHC = 180°

⇒∠FGH + ∠HEF = 180° [Vertically opposite angles]

We know that opposite angles of a cyclic quadrilateral are supplementary.

∴ EFGH is also a cyclic quadrilateral.

Hence proved

We know that Length of chord,

Where r = radius of circle, d = distance of chord from centre

⇒ l = 2 (8)

∴ Length of chord, l = 16 cm

We know that Length of chord,

Where r = radius of circle, d = distance of chord from centre

⇒

⇒

Squaring on both sides,

⇒

⇒ 576 = 4 (169 – d²)

⇒576/ 4 = 169 – d²

⇒ 144 = 169 – d²

⇒ d² = 169 – 144

⇒ d² = 25

∴ Distance of chord, d = 5 cm

A minor arc is an arc less than a semicircle.

Angle of a semicircle is 180°.

∴ Degree measure of minor arc is less than 180°.

A major arc is an arc greater than a semicircle.

Angle of a semicircle is 180°.

∴ Degree measure of major arc is greater than 180°.

From the centre to a point on the circle, it is called a radius.

Radius of a circle has constant distance.

So, the chords lying at equal distances from the centre in a circle are equal to each other.

The angle of a semicircle is 180°.

∴ The arc is a semicircle if its degree measure is 180°.

We know that there is one and only one circle passing through three non-collinear points.

A circle can be divided into two equal parts i.e. semicircles.

So, if arc AB = arc BA, then arc is a semicircle.

AB and CD are two chords.

Let EF be the diameter of the circle.

We know that if a radius bisects a chord, then it is perpendicular to the chord.

⇒ EF ⊥ CD and EF ⊥ AB

∴ AB || CD

∴ If a diameter bisects each of the chords, then the chords will be parallel.

Let us take a circle with O as centre and radius r in which arc AB ≅ arc CD.

Construction:

Join OA, OB, OC and OD.

In ΔAOB and ΔCOD,

⇒ OA = OC [radii of same circle]

⇒OB = OD [radii of same circle]

⇒∠AOB = ∠COD [measure (arc AB) = measure (arc CD)]

By SAS congruency,

⇒ ΔAOB ≅ ΔCOD

By CPCT,

⇒ AB = CD

∴ In congruent circles, if two arcs are equal, then their corresponding chords will be equal.

We know that Length of chord,

Where r = radius of circle, d = distance of chord from centre

Given diameter AD = 34 cm and chord AB = 30 cm

⇒ Radius = 34/ 2 = 17 cm

Now,

⇒

⇒

Squaring on both sides,

⇒

⇒ 900 = 4 (289 – d²)

⇒900/ 4 = 289 – d²

⇒225 = 289 – d²

⇒ d² = 289 – 225

⇒ d² = 64

∴ Distance of chord, d = 8 cm

Given OA = 5 cm, AB = 8 cm and OD ⊥ AB.

We know that the perpendicular from centre of a circle to a chord bisects the chord.

⇒ AC = CB = 1/2 AB = 1/2 (8) = 4 cm

By Pythagoras Theorem,

⇒ AO² = AC² + OC²

⇒ 5² = 4² + OC²

⇒ 25 = 16 + OC²

⇒ OC² = 25 – 16 = 9

⇒ OC = 3 cm

⇒ OA = OD = 5 cm [Radii of same circle]

⇒ CD = OD – OC = 5 – 3

∴ CD = 2 cm

We know that Length of chord,

Where r = radius of circle, d = distance of chord from centre

Given AB = 12 cm and BC = 16 cm

Now,

⇒

⇒

Squaring on both sides,

⇒

⇒ 256 = 4 (r² – 144)

⇒ 256/ 4 = r² – 144

⇒ 64 = r² – 144

⇒ d² = 289 – 225

⇒ d² = 64

∴ Distance of chord, d = 8 cm

Given ∠ABC = 20°

We know that angle subtended at the centre by an arc is twice the angle subtended by it at the remaining part of the circle.

⇒∠AOC = 2 ∠ABC

⇒∠AOC = 2 (20°)

∴ ∠AOC = 40°

Given AOB is a diameter of a circle and AC = BC.

We know that diameter subtends a right angle to the circle.

∴ ∠BCA = 90°

We know that angles opposite to equal sides are equal.

∴ ∠ABC = ∠CAB

In ΔABC,

By angle sum property,

⇒∠CAB + ∠ABC + ∠BCA = 180°

⇒∠CAB + ∠CAB + 90° = 180°

⇒ 2 ∠CAB + 90° = 180°

⇒ 2 ∠CAB = 180° - 90°

⇒∠CAB = 90°/ 2

∴ ∠CAB = 45°

Given ∠OAB = 40°

Consider ΔOAB,

⇒ OA = OB [radii of same circle]

We know that angles opposite to equal sides are equal.

⇒∠OBA = ∠OAB = 40°

By angle sum property,

⇒∠AOB + ∠OBA + ∠BAO = 180°

⇒∠AOB + 40° + 40° = 180°

⇒∠AOB = 180° - 80°

∴ ∠AOB = 100°

We know that the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

⇒∠AOB = 2 ∠ACB

⇒ 100° = 2 ∠ACB

⇒∠ACB = 100°/ 2

∴ ∠ACB = 50°

Given ∠DAB = 60° and ∠ABD = 50°

We know that angles in the same segment of a circle are equal.

∴ ∠ADB = ∠ACB

In ΔABD,

By angle sum property,

⇒∠ABD + ∠ADB + ∠DAB = 180°

⇒ 50° + ∠ADB + 60° = 180°

⇒ 110° + ∠ADB = 180°

⇒∠ADB = 180° - 110°

∴ ∠ADB = 70° = ∠ACB

Given AB of a quadrilateral is a diameter of its circumscribed circle.

And ∠ADC = 140°

Here, O is the centre of the circle and ADCBOA is the semi circle.

∴ ∠ADB = 90°

Given that ∠ADC = 140°

⇒∠ADB + ∠BDC = 140°

⇒ 90° + ∠BDC = 140°

⇒∠BDC = 140° - 90°

∴ ∠BDC = 50°

We know that angles in the same segment are equal.

⇒∠BDC = ∠BAC

∴ ∠BAC = 50°

Given BC is a diameter of the circle and ∠BAO = 60°.

Consider ΔOAB,

⇒ OA = OB [Radii of same circle]

We know that angles opposite to equal sides are equal.

⇒∠OBA = ∠BAO = 60°

By angle sum property,

⇒∠OBA + ∠BAO + ∠AOB = 180°

⇒ 60° + 60° + ∠AOB = 180°

⇒∠AOB = 180° - 120°

∴ ∠AOB = 60°

By linear pair axiom,

⇒∠AOC = 180° - 60° = 120°

We know that angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the circle.

⇒∠AOC = 2 ∠ADC

⇒ 120° = 2 ∠ADC

⇒∠ADC = 120°/ 2

∴ ∠ADC = 60°

Given ∠AOB = 90° and ∠ABC = 30°

We know that angles opposite to equal sides are equal.

⇒∠OAB = ∠ABO

By angle sum property,

⇒∠OAB + ∠ABO + ∠BOA = 180°

⇒ 2 ∠OAB + ∠BOA = 180°

⇒ 2 ∠OAB + 90° = 180°

⇒ 2 ∠OAB = 180° - 90° = 90°

⇒∠OAB = 90°/ 2

∴ ∠OAB = 45°

We know that angles subtended by an arc at the centre of the circle is double the angle subtended by it to any other part of the circle.

∴ ∠C = 45°

By angle sum property,

⇒∠ABC + ∠BCA + ∠ACB = 180°

⇒ 30° + 45° + ∠CAB = 180°

⇒ 75° + ∠CAB = 180°

⇒∠CAB = 180° - 75°

∴ ∠CAB = 105°

Also ∠CAB = ∠CAO + ∠OAB

⇒ 105° = ∠CAO + 45°

⇒∠CAO = 105° - 45°

∴ ∠CAO = 60°

Let AB and CD be two equal chords of a circle which are intersecting at a point E.

Construction:

Draw perpendiculars OF and OG on the chords.

Join OE.

Consider ΔOFE and ΔOGE,

⇒ OF = OG [Equal chords]

⇒∠OFE = ∠OGE [Each 90°]

⇒ OE = OE [Common]

By RHS congruence rule,

⇒ ΔOFE ≅ ΔOGE

By CPCT,

⇒ FE = GE … (1)

Given AB = CD … (2)

⇒ 1/2 AB = 1/2 CD

⇒ AG = CF … (3)

Adding equations (1) and (3),

⇒ AG + GE = CF + FE

⇒ AE = CE … (4)

Subtracting equation (4) from (2),

⇒ AB – AE = CD - CE

⇒ BE = DE … (5)

From (4) and (5),

We can see that two parts of one chord are equal to two parts of another chord.

Hence proved.

Given that in ΔABC, P, Q and R are the mid-points of sides BC, CA and AB. Also AD ⊥ BC.

We have to prove that P, Q, R and D are concyclic.

Proof:

In ΔABC, R and Q are mid-points of AB and CA respectively.

By mid-point theorem, RQ || BC.

Also, PQ || AB and PR || CA

Consider quadrilateral BPQR,

⇒ BP ||RQ and PQ || BR

∴ BPQR quadrilateral is a parallelogram.

Similarly, ARPQ quadrilateral is a parallelogram.

We know that opposite angles of a parallelogram are equal.

⇒∠A = ∠RPQ

PR || AC and PC is the traversal,

⇒∠BPR = ∠C

⇒∠DPQ = ∠DPR + ∠RPQ = ∠A + ∠C … (1)

RQ || BC and BR is the traversal,

⇒∠ARO = ∠B … (2)

In ΔABD, R is the mid-point of AB and OR || BD.

∴ O is the mid-point of AD.

⇒ OA = OD

In ΔAOR and ΔDOR,

⇒ OA = OD

⇒∠AOR = ∠DOR = 90°

⇒ OR = OR [Common]

By SAS congruence rule,

⇒ ΔAOR ≅ Δ DRO

⇒∠ARO = ∠DRO [CPCT]

⇒∠DRO = ∠B [From (2)]

In quadrilateral PRQD,

⇒∠DRO + ∠DPQ = ∠B + (∠A + ∠C) = ∠A + ∠B + ∠C [From (1)]

Since ∠A + ∠B + ∠C = 180°,

⇒∠DRO + ∠DPQ = 180°

Hence, quadrilateral PRQD is a cyclic quadrilateral.

∴ Points P, Q, R and D are concyclic.

Hence proved.

Given ABCD is a parallelogram.

A circle through points A and B is drawn such that it intersects the AD at P and BC at Q.

We have to prove that points P, Q, C and D are cyclic.

Proof:

Since the circle passes through points A, B, P and Q, ABPQ is a cyclic quadrilateral.

We know that opposite angles in a cyclic quadrilateral are supplementary.

⇒∠A + ∠PQB = 180°

⇒∠CQP + ∠PQB = 180°

∴ ∠A = ∠CQP

Now, AB and CD are parallel lines and AD is the traversal.

We know that angles on the same side of traversal are supplementary.

⇒∠A + ∠D = 180°

⇒∠CQP + ∠D = 180°

Thus, in PQCD quadrilateral, opposite angles are supplementary.

Hence, quadrilateral PQCD is a cyclic quadrilateral.

∴ P, Q, C and D are concyclic.

Hence proved.

Here, O is the circumcentre of ΔABC.

Let the bisector AD of ∠A and perpendicular bisector OD of BC intersect at D.

[here perpendicular bisector passes through center because circumcenter of any triangle lies on perpendicular bisector of any of its side]

Proof:

We know that angle subtended by an arc at the centre is twice the angle subtended by the arc at the point of the alternate segment of the circle.

⇒∠BOC = 2∠A

⇒ OB = OC [Radii of same circle]

∴ ΔBOC is an isosceles triangle.

Since OD is the perpendicular bisector of BC,

We know that any three points arc always concyclic.

∴ A, C and D are concyclic.

As, AD is angle bisector,

⇒ Arc CD subtends at point A and at point O

∴ O is the centre of the circle passing through A, C and D.

Thus, the circle passing through A, C and D is the circum circle of ΔABC.

⇒ D passes through circumcircle of ΔABC

∴ They intersect the circum circle of ΔABC.

Hence proved.

Given AB and CD are two chords of a circle AYDZBWCX such that AB ⊥ CD.

Let the chords intersect at O.

We have to prove that arc CXA + arc DBZ = arc AYD + arc CWB = semicircle.

Construction:

Join AD, DB, AC and BC.

Proof:

Angle subtended by chord AC is ∠CBA and angle subtended by chord BD is ∠BCD.

⇒∠CBA + ∠BCD = 180° - ∠COB

= 180° - 90°

= 90°

Since angle subtended in a semicircle is 90°,

⇒ arc CXA + arc DBZ = semicircle

Similarly arc AYD + arc CWB = semicircle

∴ arc CXA + arc DBZ = arc AYD + arc CWB = semicircle

Hence proved.

Given ΔABC is an equilateral triangle inscribed in a circle with centre O.

P is a point lying on minor arc BC.

We have to prove that PA is bisector of ∠BPC.

Construction:

Join OA, OB, OC, BP and PC.

Proof:

Since ΔABC is equilateral,

⇒ AB = BC = CA

We know that equal chords subtend equal angles at centre.

⇒∠AOB = ∠AOC = ∠BOC

Consider ∠AOB = ∠AOC … (1)

∠AOB and ∠APB are angles subtended by an arc AB at centre and at remaining part of the circle by same arc.

⇒∠APB = 1/2 ∠AOB … (2)

⇒∠APC = 1/2 ∠AOC … (3)

From (1), (2) and (3),

∠APB = ∠APC

∴ PA is angle bisector of ∠BPC.

Hence proved.

Given AB and CD are two chords of the circle with centre O which intersects at E.

We have to prove that ∠AEC = 1/2 (angle subtended by arc CXA at centre + angle subtended by arc DYB at centre).

Construction:

Join AC, BC and BD.

Proof:

AC is a chord.

We know that angle subtended at center is double the angle subtended at circumference.

⇒∠AOC = 2∠ABC … (1)

⇒∠DOB = 2∠DCB … (2)

Adding (1) and (2),

⇒∠AOC + ∠DOB = 2(∠ABC + ∠DCB) … (3)

Consider ΔCEB,

We know that the sum of two opposite interior angles is the exterior angle.

⇒∠AEC = ∠ECB + ∠CBE

⇒∠AEC = ∠DCB + ∠ABC … (4)

From (3) and (4),

∴ ∠AOC + ∠DOB = 2 (∠AEC)

Hence proved.

Given ABCD is a cyclic quadrilateral. AP and CQ are bisectors of ∠A and ∠C respectively.

We have to prove that PQ is the diameter of the circle.

Construction:

Join AF and FD

Proof:

We know that in a cyclic quadrilateral, opposite angles are supplementary.

⇒∠A + ∠C = 180°

⇒ 1/2 ∠A + 1/2 ∠C = 90°

⇒∠EAD + ∠DCF = 90° … (1)

We know that angles in the same segment are equal.

⇒∠DCF = ∠DAF … (2)

From (1) and (2),

⇒∠EAD + ∠DAF = 90°

⇒∠EAF = 90°

∠EAF is the angle in a semicircle.

∴ EF is the diameter of the circle.

Hence proved.

Given radius of circle OQ = OR = √2 cm

Length of chord, QR = 2 cm

In ΔOQR,

⇒ OQ² + OR² = (√2)² + (√2)²

= 2 + 2

= 4

= (QR)²

⇒ OQ² + OR² = QR²

∴ ΔOQR is a right angled triangle at ∠QOR.

We know that angle at the centre is double the angle on the remaining part of the circle.

⇒∠QOR = 2∠QPR

⇒ 90° = 2∠QPR

∴ ∠QPR = 45°

∴ The angle subtended by the chord at the point in major segment is 45°.

Hence proved

Given AB and AC are two chords of a circle of radius r such that AB = 2AC.

P and q are perpendicular distances of AB and AC from centre O.

⇒ OM = p and ON = q

We have to prove that 4q² = p² + 3r²

Proof:

We know that perpendicular from centre to chord intersect at mid-point of the chord.

Consider ΔONA,

By Pythagoras Theorem,

⇒ ON² + NA² = OA²

⇒ q² + NA² = r²

⇒ NA² = r² – q² … (1)

Consider ΔOMA,

By Pythagoras Theorem,

⇒ OM² + AM² = OA²

⇒ p² + AM² = r²

⇒ AM² = r² – p² … (2)

⇒ AM = = = AC = 2NA

From (1) and (2),

⇒ r² – p² = AM² = (2NA)² = 4NA²

⇒ r² – p² = 4 (r² – q²)

⇒ r² – p² = 4r² – 4q²

∴ 4q² = 3r² + p²

Hence proved.

Given O is the centre of the circle and ∠BCO = 30°.

Construction:

Join AC and OB.

In ΔOBC,

⇒ OC = OB [Radii of circle]

⇒∠OBC = ∠OCB = 30°

From angle sum property,

⇒∠OBC + ∠OCB + ∠BOC = 180°

⇒ 30° +30° + ∠BOC = 180°

⇒∠BOC = 180° - 60°

∴ ∠BOC = 120°

We know that the central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.

⇒∠BOC = 2∠BAC

⇒ 2∠BAC = 120°

∴ ∠BAC = 60°

⇒∠AEB = 90°

So, OE ⊥ BC

We know that a line form center to any chord is perpendicular then that line also bisects chord.

⇒ CE = BE … (1)

Consider ΔABE and ΔACE,

⇒ CE = BE [From (1)]

⇒∠AEB = ∠AEC = 90°

⇒ AE = AE [Common side]

By RHL rule,

⇒ ΔABE ≅ ΔACE

By CPCT,

∠BAE = ∠CAE

⇒∠BAC = ∠BAE + ∠CAE

⇒ x + x = 60°

⇒ 2x = 60°

∴ x = 30°

And ∠COI = ∠OCB = 30° [Alternate interior angles]

We know that the central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.

⇒∠COI = 2∠CBI

⇒ 2y = 30°

∴ y = 15°

Given O is the centre of the circle, BD = OD and CD ⊥ AB.

⇒ OD = OB [Radii of circle]

⇒In ΔOBD,

⇒ OD = BD = OB

∴ ΔOBD is an equilateral triangle.

Each interior angle of an equilateral triangle is 60°.

⇒∠OBD = ∠ODB = ∠BOD = 60°

Also given CD ⊥ AB,

We know that in equilateral triangle altitude drawn from any vertex bisects the vertex angle and also bisects the opposite side.

⇒∠EDB = ∠EDO = 30° … (1) and OE = BE

We know that if a perpendicular is drawn from center to any chord, it bisects the chord.

⇒ CE = DE … (2)

Consider ΔBED and ΔBEC,

⇒ CE = DE [From equation (2)]

⇒∠BED = ∠BEC = 90° [CD ⊥ AB]

⇒ BE = BE [Common side]

By SAS rule,

⇒ ΔBED ≅ ΔBEC

By CPCT,

⇒∠BDE = ∠BCE

From (1),

⇒∠BDE = ∠BCE = 30°

In ΔBCE,

From angle sum property,

⇒∠BCE + ∠BEC + ∠CBE = 180°

⇒ 30° + 90° + ∠CBE = 180°

⇒∠CBE = 180° - 120°

∴ ∠CBE = 60°

We know that the angle inscribed in a semicircle is always a right angle.

In ΔABC,

From angle sum property,

⇒∠CBA + ∠ACB + ∠CAB = 180°

⇒ 60° + 90° + ∠CAB = 180° [∠CBA = ∠CBE]

⇒∠CAB = 180° - 150°

∴ ∠CAB = 30°

Let P be the given point inside a circle with centre O.

Draw the chord AB which is perpendicular to the diameter XY through P.

Draw ON ⊥ CD from O.

Then ΔONP is a right angled triangle.

⇒Its hypotenuse OP is larger than ON.

We know that the chord nearer to the centre is larger than the chord which is farther to the centre.

⇒ CD > AB

∴ AB is the smallest of all chords passing through P.