NCERT Solutions Class 9 Mathematics Chapter 10 Circles

Exercise 10.1

Q.1) Fill in the blanks :
(i) The centre of a circle lies in ___________ of the circle. (exterior/interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in ____ of the circle. (exterior/interior)
(iii) The longest chord of a circle is a __________ of the circle.
(iv) An arc is a __________ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and __________ of the circle.
(vi) A circle divides the plane, on which it lies in __________ parts.
Sol.1) (i) interior (ii) exterior (iii) diameter (iv) semicircle
(v) the chord (vi) three

Q.2) Write True or False: Give reasons for your answers.
(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.
Sol.2) (i) True.
All the line segment from the centre to the circle is of equal length.
(ii) False.
We can draw infinite numbers of equal chords.
(iii) False.
We get major and minor arcs for unequal arcs. So, for equal arcs on circle we can’t say it is major arc or minor arc.
(iv) True.
A chord which is twice as long as radius must pass through the centre of the circle and is diameter to the circle.
(v) False.
Sector is the region between the arc and the two radii of the circle.
(vi) True.
A circle can be drawn on the plane.

Exercise 10.2

Q.1) Recall that two circles are congruent if they have the same radii. Prove that equal chords
of congruent circles subtend equal angles at their centres.
Sol.1) Given : Two congruent circles with centres O and 𝑂′. AB and CD are equal chords of the circles with centres O and 𝑂′ respectively.
To Prove : ∠𝐴𝑂𝐵 = ∠𝐶𝑂𝐷
Proof : In triangles AOB and COD,
AB = CD [Given]
AO = CO [Radii of congruent circle]
BO = DO [Radii of congruent circle]
⇒ Δ𝐴𝑂𝐵 ≅ Δ𝐶𝑂′𝐷 [SSS axiom]

⇒ ∠𝐴𝑂𝐵 ≅ ∠𝐶𝑂′𝐷 [CPCT] Proved.

Q.2) Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Sol.2) Given : Two congruent circles with centres O and O′. AB and CD are chords of circles with centre O and O′ respectively such that ∠𝐴𝑂𝐵 = ∠𝐶𝑂′𝐷
To Prove : AB = CD
Proof : In triangles 𝐴𝑂𝐵 and 𝐶𝑂′𝐷,
𝐴𝑂 = 𝐶𝑂 [Radii of congruent circle]
𝐵𝑂 = 𝐷𝑂 [Radii of congruent circle]
∠𝐴𝑂𝐵 = ∠𝐶𝑂′𝐷 [Given]
⇒ Δ𝐴𝑂𝐵 ≅ Δ𝐶𝑂′𝐷 [SAS axiom]
⇒ 𝐴𝐵 = 𝐶𝐷 [CPCT] Proved.

Exercise 10.3

Q.1) Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Sol.1)

Maximum number of common points = 2

Q.2) Suppose you are given a circle. Give a construction to find its centre.
Sol.2) Steps of Construction:

1. Take arc PQ of the given circle.
2. Take a point R on the arc PQ and draw chords PR and RQ.
3. Draw perpendicular bisectors of PR and RQ`.
These perpendicular bisectors intersect at point O.
Hence, point O is the centre of the given circle.

Q.3) If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Sol.3) Given : AB is the common chord of two intersecting circles (𝑂, 𝑟) and (𝑂′, 𝑟′).
To Prove : Centres of both circles lie on the perpendicular bisector of chord AB, i.e., AB is bisected at right angle by 𝑂𝑂′.
Construction : Join 𝐴𝑂, 𝐵𝑂, 𝐴𝑂′ and 𝐵𝑂′.
Proof : In Δ𝐴𝑂𝑂′ and Δ𝐵𝑂𝑂′
𝐴𝑂 = 𝑂𝐵                                    (Radii of the circle (𝑂, 𝑟)
𝐴𝑂′ = 𝐵𝑂′                                  (Radii of the circle (𝑂′, 𝑟′))
𝑂𝑂′ = 𝑂𝑂′                                  (Common)
∴ Δ𝐴𝑂𝑂′ ≅ Δ𝐵𝑂𝑂′                     (SSS congruency)
⇒ ∠𝐴𝑂𝑂′ = ∠𝐵𝑂𝑂                     (CPCT)
Now in Δ𝐴𝑂𝐶 and Δ𝐵𝑂𝐶
∠𝐴𝑂𝐶 = ∠𝐵𝑂𝐶 (∠𝐴𝑂𝑂′ = ∠𝐵𝑂𝑂′)
𝐴𝑂 = 𝐵𝑂 (Radii of the circle (O, r))
𝑂𝐶 = 𝑂𝐶 (Common)
∴ Δ𝐴𝑂𝐶 ≅ Δ𝐵𝑂𝐶 (SAS congruency)
⇒ 𝐴𝐶 = 𝐵𝐶 and ∠𝐴𝐶𝑂 = ∠𝐵𝐶𝑂         ...(i) (CPCT)
⇒ ∠𝐴𝐶𝑂 + ∠𝐵𝐶𝑂 = 180°                     (ii) (Linear pair)
⇒ ∠𝐴𝐶𝑂 + ∠𝐵𝐶𝑂 = 90°                      [from (i) & (ii)]

Hence, 𝑂𝑂′ lie on the perpendicular bisector of AB.

Exercise

10.4

Q.1) Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Sol.1) In Δ𝐴𝑂𝑂′,
𝐴𝑂2 = 52 = 25
𝐴𝑂′2 = 32 = 9
𝑂𝑂′2 = 42 = 16
𝐴𝑂′2 + 𝑂𝑂′2 = 9 + 16 = 25 = 𝐴𝑂2
⇒ ∠𝐴𝑂′𝑂 = 90° [By converse of Pythagoras theorem]
Similarly, ∠𝐵𝑂′𝑂 = 90°.
⇒ ∠𝐴𝑂′𝐵 = 90° + 90° = 180°
⇒ 𝐴𝑂′𝐵 is a straight line, whose mid-point is O.
⇒ 𝐴𝐵 = (3 + 3) 𝑐𝑚 = 6 𝑐𝑚 Ans.

Q.2) If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Sol.2) Given : AB and CD are two equal chords of a circle which meet at E.
To prove : 𝐴𝐸 = 𝐶𝐸 and 𝐵𝐸 = 𝐷𝐸
Construction : Draw 𝑂𝑀 ⊥ 𝐴𝐵 and 𝑂𝑁 ⊥ 𝐶𝐷 and join OE.
Proof : In Δ𝑂𝑀𝐸 and Δ𝑂𝑁𝐸
𝑂𝑀 = 𝑂𝑁             [Equal chords are equidistant]
OE = OE             [Common]
∠𝑂𝑀𝐸 = ∠𝑂𝑁𝐸               [Each equal to 90°]
∴ Δ𝑂𝑀𝐸 ≅ Δ𝑂𝑁𝐸            [RHS axiom]
⇒ 𝐸𝑀 = 𝐸𝑁             ..(i) [CPCT]
Now 𝐴𝐵 = 𝐶𝐷            [Given]
⇒ 1/2 𝐴𝐵 = 1/2 𝐶𝐷
⇒ 𝐴𝑀 = 𝐶𝑁 ..(ii) [Perpendicular from centre bisects the chord]
Adding (i) and (ii), we get
𝐸𝑀 + 𝐴𝑀 = 𝐸𝑁 + 𝐶𝑁
⇒ 𝐴𝐸 = 𝐶𝐸 ..(iii)
Now, 𝐴𝐵 = 𝐶𝐷 ..(iv)
⇒ 𝐴𝐵 – 𝐴𝐸 = 𝐶𝐷 – 𝐴𝐸 [From (iii)]
⇒ 𝐵𝐸 = 𝐶𝐷 – 𝐶𝐸 Proved.

Q.3) If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Sol.3) Given : AB and CD are two equal chords of a circle which meet at E within the circle and
a line PQ joining the point of intersection to the centre.
To Prove : ∠𝐴𝐸𝑄 = ∠𝐷𝐸𝑄
Construction : Draw 𝑂𝐿 ⊥ 𝐴𝐵 and 𝑂𝑀 ⊥ 𝐶𝐷.
Proof : In Δ𝑂𝐿𝐸 and Δ𝑂𝑀𝐸, we have
𝑂𝐿 = 𝑂𝑀             [Equal chords are equidistant]
𝑂𝐸 = 𝑂𝐸             [Common]
∠𝑂𝐿𝐸 = ∠𝑂𝑀𝐸               [Each = 90°]
∴ Δ𝑂𝐿𝐸 ≅ Δ𝑂𝑀𝐸            [RHS congruence]
⇒ ∠𝐿𝐸𝑂 = ∠𝑀𝐸𝑂           [CPCT]

Q.4) If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig.)
Sol.4) Given : A line AD intersects two concentric circles at A, B, C and D, where O is the centre of these circles.
To prove : 𝐴𝐵 = 𝐶𝐷
Construction : Draw 𝑂𝑀 ⊥ 𝐴𝐷.
Proof : AD is the chord of larger circle.
∴ 𝐴𝑀 = 𝐷𝑀 ..                       (i)                       [OM bisects the chord]
BC is the chord of smaller circle
∴ 𝐵𝑀 = 𝐶𝑀 ..                       (ii)                       [OM bisects the chord]
Subtracting                         (ii) from (i), we get
𝐴𝑀 – 𝐵𝑀 = 𝐷𝑀 – 𝐶𝑀
⇒ 𝐴𝐵 = 𝐶𝐷                        Proved.

Q.5) Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Sol.5) Let Reshma, Salma and Mandip be represented by R, S and M respectively.
Draw 𝑂𝐿 ⊥ 𝑅𝑆,
𝑂𝐿² = 𝑂𝑅²– 𝑅𝐿²
𝑂𝐿² = 52 – 32 [𝑅𝐿 = 3 𝑚, because 𝑂𝐿 ⊥ 𝑅𝑆]
= 25 – 9 = 16
𝑂𝐿 = √16 = 4
Now, area of triangle 𝑂𝑅𝑆 = 1/2 × 𝐾𝑅 × 05
= 1/2 × 𝐾𝑅 × 05
Also, area of Δ𝑂𝑅𝑆 = 1/2 × 𝑅𝑆 × 𝑂𝐿
= 1/2× 6 × 4 = 12 𝑚²
⇒ 1/2 × 𝐾𝑅 × 5 = 1/2
⇒ 𝐾𝑅 = 12 × 2/5 = 24/5
= 4.8 𝑚
⇒ 𝑅𝑀 = 2𝐾𝑅
⇒ 𝑅𝑀 = 2 × 4.8 = 9.6 𝑚
Hence, distance between Reshma and Mandip is 9.6 𝑚 Ans.

Q.6) A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are siting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
Sol.6) Let Ankur, Syed and David be represented by A, S and D respectively.
Let 𝑃𝐷 = 𝑆𝑃 = 𝑆𝑄 = 𝑄𝐴 = 𝐴𝑅 = 𝑅𝐷 = 𝑥
In Δ𝑂𝑃𝐷,
𝑂𝑃² = 400 – 𝑥²
⇒ 𝑂𝑃 = 400 − 𝑥²

⇒ 𝑥² = 4𝑥² – 3600 + 9𝑥²
⇒ 12𝑥² = 3600
⇒ 𝑥² = 3600
12 = 300
⇒ 𝑥 = 10√3
Now, 𝑆𝐷 = 2𝑥 = 2 × 10 √3 = 20 √3
∴ 𝐴𝑆𝐷 is an equilateral triangle.
⇒ 𝑆𝐷 = 𝐴𝑆 = 𝐴𝐷 = 20 √3
Hence, length of the string of each phone is 20 √3𝑚 Ans.

Exercise 10.5

Q.1) In the figure, A, B and C are three points on a circle with centre O such that ∠ 𝐵𝑂𝐶 = 30° and ∠ 𝐴𝑂𝐵 = 60°. If D is a point on the circle other than the arc ABC, find ∠ 𝐴𝐷𝐶.
Sol.1) We have, ∠𝐵𝑂𝐶 = 30° and ∠𝐴𝑂𝐵 = 60°
∠𝐴𝑂𝐶 = ∠𝐴𝑂𝐵 + ∠𝐵𝑂𝐶
= 60° + 30° = 90°
We know that angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc on the remaining part of the circle.
∴ 2∠𝐴𝐷𝐶 = ∠𝐴𝑂𝐶
⇒ ∠𝐴𝐷𝐶 = 1/2
∠𝐴𝑂𝐶 = 1/2 × 90°
⇒ ∠𝐴𝐷𝐶 = 45°

Q.2) A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Sol.2) We have, 𝑂𝐴 = 𝑂𝐵 = 𝐴𝐵
Therefore, Δ𝑂𝐴𝐵 is a equilateral triangle.
⇒ ∠𝐴𝑂𝐵 = 60°
We know that angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc on the remaining part of the circle.
∴ ∠𝐴𝑂𝐵 = 2∠𝐴𝐶𝐵
⇒ ∠𝐴𝐶𝐵 = 1/2 ∠𝐴𝑂𝐵 = 1/2 × 60°
⇒ ∠𝐴𝐶𝐵 = 30°
Also, ∠𝐴𝐷𝐵 = 1/2 reflex ∠𝐴𝑂𝐵
= 1/2 (360° – 60°) = 1/2 × 300° = 150°
Hence, angle subtended by the chord at a point on the minor arc is 150° and at a point on the major arc is 30°

Q.3) In the figure, ∠𝑃𝑄𝑅 = 100°, where 𝑃, 𝑄 and R are points on a circle with centre O. Find ∠𝑂𝑃𝑅.
Sol.3) Reflex angle 𝑃𝑂𝑅 = 2∠𝑃𝑄𝑅
= 2 × 100° = 200°
Now, angle 𝑃𝑂𝑅 = 360° – 200 = 160°
Also, 𝑃𝑂 = 𝑂𝑅                       [Radii of a circle]
∠𝑂𝑃𝑅 = ∠𝑂𝑅𝑃                       [Opposite angles of isosceles triangle]
In Δ𝑂𝑃𝑅, ∠𝑃𝑂𝑅 = 160°
∴ ∠𝑂𝑃𝑅 = ∠𝑂𝑅𝑃 = 10°          [Angle sum property of a triangle] Ans.

Q.4) In Fig., ∠𝐴𝐵𝐶 = 69°, ∠ 𝐴𝐶𝐵 = 31°, find ∠𝐵𝐷𝐶.
Sol.4) ∠𝐵𝐴𝐶 = ∠𝐵𝐷𝐶 (Angles in the segment of the circle)
In 𝛥𝐴𝐵𝐶,
∠𝐵𝐴𝐶 + ∠𝐴𝐵𝐶 + ∠𝐴𝐶𝐵 = 180° (Sum of the angles in a triangle)
⇒ ∠𝐵𝐴𝐶 + 69° + 31° = 180°
⇒ ∠𝐵𝐴𝐶 = 180° – 100°
⇒ ∠𝐵𝐴𝐶 = 80°
Thus, ∠𝐵𝐷𝐶 = 180°

Q.5) In Fig., A, B, C and D are four points on a circle. AC and BD intersect at a point E such that
∠𝐵𝐸𝐶 = 130° and ∠𝐸𝐶𝐷 = 20°. Find ∠ 𝐵𝐴𝐶.
Sol.5) ∠𝐵𝐴𝐶 = ∠𝐶𝐷𝐸           (Angles in the segment of the circle)
In 𝛥𝐶𝐷𝐸,
∠𝐶𝐸𝐵 = ∠𝐶𝐷𝐸 + ∠𝐷𝐶𝐸         (Exterior angles of the triangle)
⇒ 130° = ∠𝐶𝐷𝐸 + 20°
⇒ ∠𝐶𝐷𝐸 = 110°
Thus, ∠𝐵𝐴𝐶 = 110°

Q.6) ABCD is a cyclic quadrilateral whose diagonals intersect at a point E.

If ∠ 𝐷𝐵𝐶 = 70°, ∠ 𝐵𝐴𝐶 = 30°, find ∠ 𝐵𝐶𝐷. Further, if 𝐴𝐵 = 𝐵𝐶, find ∠ 𝐸𝐶𝐷.
Sol.6) ∠𝐶𝐴𝐷 = ∠𝐷𝐵𝐶 = 70°                 [Angles in the same segment]
Therefore, ∠𝐷𝐴𝐵 = ∠𝐶𝐴𝐷 + ∠𝐵𝐴𝐶
= 70° + 30° = 100°
But, ∠𝐷𝐴𝐵 + ∠𝐵𝐶𝐷 = 180°                   [Opposite angles of a cyclic quadrilateral]
So, ∠𝐵𝐶𝐷 = 180° – 100° = 80°
Now, we have 𝐴𝐵 = 𝐵𝐶
Therefore, ∠𝐵𝐶𝐴 = 30°                       [Opposite angles of an isosceles triangle]
Again, ∠𝐷𝐴𝐵 + ∠𝐵𝐶𝐷 = 180°               [Opposite angles of a cyclic quadrilateral]
⇒ 100° + ∠𝐵𝐶𝐴 + ∠𝐸𝐶𝐷 = 180°           [∵ ∠𝐵𝐶𝐷 = ∠𝐵𝐶𝐴 + ∠𝐸𝐶𝐷]
⇒ 100° + 30° + ∠𝐸𝐶𝐷 = 180°
⇒ 130° + ∠𝐸𝐶𝐷 = 180°
⇒ ∠𝐸𝐶𝐷 = 180° – 130° = 50°
Hence, ∠𝐵𝐶𝐷 = 80° and ∠𝐸𝐶𝐷 = 50°

Q.7) If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Sol.7) Given : ABCD is a cyclic quadrilateral, whose diagonals AC and BD are diameter of the
circle passing through A, B, C and D.
To Prove : 𝐴𝐵𝐶𝐷 is a rectangle.
Proof : In Δ𝐴𝑂𝐷 and Δ𝐶𝑂𝐵
𝐴𝑂 = 𝐶𝑂                     [Radii of a circle]
𝑂𝐷 = 𝑂𝐵                     [Radii of a circle]
∠𝐴𝑂𝐷 = ∠𝐶𝑂𝐵              [Vertically opposite angles]
∴ Δ𝐴𝑂𝐷 ≅ Δ𝐶𝑂𝐵           [SAS axiom]
∴ ∠𝑂𝐴𝐷 = ∠𝑂𝐶𝐵           [CPCT]

But these are alternate interior angles made by the transversal AC, intersecting AD and BC.
∴ 𝐴𝐷 || 𝐵𝐶
Similarly, 𝐴𝐵 || 𝐶𝐷.
Hence, quadrilateral ABCD is a parallelogram.
Also, ∠𝐴𝐵𝐶 = ∠𝐴𝐷𝐶       ..(i) [Opposite angles of a ||𝑔𝑚 are equal]
And, ∠𝐴𝐵𝐶 + ∠𝐴𝐷𝐶 = 180° ...(ii)      [Sum of opposite angles of a cyclic quadrilateral is 180°]
⇒ ∠𝐴𝐵𝐶 = ∠𝐴𝐷𝐶 = 90°                   [From (i) and (ii)]
∴ ABCD is a rectangle.                   [A ||𝑔𝑚 one of whose angles is 90° is a rectangle]
Proved.

Q.8) If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Sol.8) Given : A trapezium ABCD in which 𝐴𝐵 || 𝐶𝐷 and 𝐴𝐷 = 𝐵𝐶.
To Prove : ABCD is a cyclic trapezium.
Construction : Draw 𝐷𝐸 ⊥ 𝐴𝐵 and 𝐶𝐹 ⊥ 𝐴𝐵.
Proof : In Δ𝐷𝐸𝐴 and Δ𝐶𝐹𝐵, we have
𝐴𝐷 = 𝐵𝐶             [Given]
∠𝐷𝐸𝐴 = ∠𝐶𝐹𝐵 = 90°             [DE ⊥ AB and CF ⊥ AB]
𝐷𝐸 = 𝐶𝐹                              [Distance between parallel lines remains constant]
∴ Δ𝐷𝐸𝐴 ≅ Δ𝐶𝐹𝐵                                [RHS axiom]
⇒ ∠𝐴 = ∠𝐵                         ...(i)       [CPCT]
and, ∠𝐴𝐷𝐸 = ∠𝐵𝐶𝐹                   ..(ii)  [CPCT]
Since, ∠𝐴𝐷𝐸 = ∠𝐵𝐶𝐹                        [From (ii)]
⇒ ∠𝐴𝐷𝐸 + 90° = ∠𝐵𝐶𝐹 + 90°
⇒ ∠𝐴𝐷𝐸 + ∠𝐶𝐷𝐸 = ∠𝐵𝐶𝐹 + ∠𝐷𝐶𝐹
⇒ ∠𝐷 = ∠𝐶 ..(iii)
[∠𝐴𝐷𝐸 + ∠𝐶𝐷𝐸 = ∠𝐷, ∠𝐵𝐶𝐹 + ∠𝐷𝐶𝐹 = ∠𝐶]
∴ ∠𝐴 = ∠𝐵 𝑎𝑛𝑑 ∠𝐶 = ∠𝐷                 [From (i) and (iii)] ……(iv)
∠𝐴 + ∠𝐵 + ∠𝐶 + ∠𝐷 = 360°           [Sum of the angles of a quadrilateral is 360°]
⇒ 2(∠𝐵 + ∠𝐷) = 360°                   [Using (iv)]
⇒ ∠𝐵 + ∠𝐷 = 180°
⇒ Sum of a pair of opposite angles of quadrilateral ABCD is 180°.
⇒ ABCD is a cyclic trapezium Proved.

Q.9) Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig.). Prove that ∠ 𝐴𝐶𝑃 = ∠𝑄𝐶𝐷.

Sol.9) Given : Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively.
To Prove : ∠𝐴𝐶𝑃 = ∠𝑄𝐶𝐷.
Proof : ∠𝐴𝐶𝑃 = ∠𝐴𝐵𝑃          ...(i) [Angles in the same segment]
∠𝑄𝐶𝐷 = ∠𝑄𝐵𝐷                    ..(ii) [Angles in the same segment]
But, ∠𝐴𝐵𝑃 = ∠𝑄𝐵𝐷             ..(iii) [Vertically opposite angles]
By (i), (ii) and (ii) we get
∠𝐴𝐶𝑃 = ∠𝑄𝐶𝐷 Proved.

Q.10) If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Sol.10) Given : Sides AB and AC of a triangle ABC are diameters of two circles which intersect at D.
To Prove : D lies on BC.
Proof : Join AD
∠𝐴𝐷𝐵 = 90° ...(i) [Angle in a semicircle]
Also, ∠𝐴𝐷𝐶 = 90° ..(ii)
Adding (i) and (ii), we get
∠𝐴𝐷𝐵 + ∠𝐴𝐷𝐶 = 90° + 90°
⇒ ∠𝐴𝐷𝐵 + ∠𝐴𝐷𝐶 = 180°
⇒ 𝐵𝐷𝐶 is a straight line.
∴ 𝐷 lies on 𝐵𝐶

Hence, point of intersection of circles lie on the third side BC. Proved.

Q.11) ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠𝐶𝐴𝐷 = ∠𝐶𝐵𝐷.
Sol.11) Given : ABC and ADC are two right triangles with common hypotenuse AC.
To Prove : ∠𝐶𝐴𝐷 = ∠𝐶𝐵𝐷
Proof : Let O be the mid-point of AC.
Then 𝑂𝐴 = 𝑂𝐵 = 𝑂𝐶 = 𝑂𝐷
Mid-point of the hypotenuse of a right triangle is equidistant from its vertices with O as centre and radius equal to OA, draw a circle to pass through A, B, C and D.
We know that angles in the same segment of a circle are equal.
Since, ∠𝐶𝐴𝐷 and ∠𝐶𝐵𝐷 are angles of the same segment.
Therefore, ∠𝐶𝐴𝐷 = ∠𝐶𝐵𝐷. Proved.

Q.12) Prove that a cyclic parallelogram is a rectangle.

Sol.12) Given : ABCD is a cyclic parallelogram.
To prove : ABCD is a rectangle.
Proof : ∠𝐴𝐵𝐶 = ∠𝐴𝐷𝐶 ...(i) [Opposite angles of a ||𝑔𝑚 are equal]
But, ∠𝐴𝐵𝐶 + ∠𝐴𝐷𝐶 = 180°... (ii)[Sum of opposite angles of a cyclic quadrilateral is 180°]
⇒ ∠𝐴𝐵𝐶 = ∠𝐴𝐷𝐶 = 90° [From (i) and (ii)]
∴ 𝐴𝐵𝐶𝐷 is a rectangle [A ||𝑔𝑚 one of whose angles is 90° is a rectangle]
Hence, a cyclic parallelogram is a rectangle. Proved.

Exercise 10.6

Q.1) Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Sol.1) Given : Two intersecting circles, in which OO′ is the line of centres and A and B are two points of intersection.
To prove : ∠𝑂𝐴𝑂′ = ∠𝑂𝐵𝑂′
Construction : Join 𝐴𝑂, 𝐵𝑂, 𝐴𝑂′ and 𝐵𝑂′.
Proof : In Δ𝐴𝑂𝑂′ and Δ𝐵𝑂𝑂′, we have
𝐴𝑂 = 𝐵𝑂      [Radii of the same circle]
𝐴𝑂′ = 𝐵𝑂′    [Radii of the same circle]
𝑂𝑂′ = 𝑂𝑂′    [Common]
∴ Δ𝐴𝑂𝑂′ ≅ Δ𝐵𝑂𝑂′ [SSS axiom]
⇒ ∠𝑂𝐴𝑂′ = ∠𝑂𝐵𝑂′ [CPCT]
Hence, the line of centres of two intersecting circles subtends equal angles at the two points of intersection. Proved.

Q.2) Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Sol.2) Let O be the centre of the circle and let its radius be 𝑟 𝑐𝑚.
Draw 𝑂𝑀 ⊥ 𝐴𝐵 and 𝑂𝐿 ⊥ 𝐶𝐷.
Then, 𝐴𝑀 = 1/2
𝐴𝐵 = 5/2 𝑐𝑚
and, 𝐶𝐿 = 1/2
𝐶𝐷 = 11/2 𝑐𝑚
Since, 𝐴𝐵 || 𝐶𝐷, it follows that the points O, L, M are collinear and therefore, 𝐿𝑀 = 6 𝑐𝑚.
Let 𝑂𝐿 = 𝑥 𝑐𝑚. Then 𝑂𝑀 = (6 – 𝑥) 𝑐𝑚
Join 𝑂𝐴 and 𝑂𝐶. Then 𝑂𝐴 = 𝑂𝐶 = 𝑟 𝑐𝑚.
Now, from right-angled Δ𝑂𝑀𝐴 and Δ𝑂𝐿𝐶, we have
𝑂𝐴² = 𝑂𝑀² + 𝐴𝑀² and 𝑂𝐶² = 𝑂𝐿² + 𝐶𝐿²                 [By Pythagoras Theorem]
⇒ 𝑟² = (6 – 𝑥)² + (5/2)²            ..(i) and 𝑟² = 𝑥² + (11/2)²                 ... (ii)
⇒ (6 – 𝑥)² + (5/2)² = 𝑥² + (11/2)²                       [From (i) and (ii)]

Q.3) The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
Sol.3) Let PQ and RS be two parallel chords of a circle with centre O.
We have, 𝑃𝑄 = 8 𝑐𝑚 and 𝑅𝑆 = 6 𝑐𝑚.
Draw perpendicular bisector OL of RS which meets 𝑃𝑄 in M. Since, 𝑃𝑄 || 𝑅𝑆, therefore,
OM is also perpendicular bisector of PQ.
Also, 𝑂𝐿 = 4 𝑐𝑚 and 𝑅𝐿 = 1/2
𝑅𝑆 ⇒ 𝑅𝐿 = 3 𝑐𝑚
and 𝑃𝑀 = 1/2
𝑃𝑄 ⇒ 𝑃𝑀 = 4 𝑐𝑚
In Δ𝑂𝑅𝐿, we have
𝑂𝑅² = 𝑅𝐿² + 𝑂𝐿²                                [Pythagoras theorem]
⇒ 𝑂𝑅² = 3² + 4² = 9 + 16
⇒ 𝑂𝑅2 = 25 ⇒ 𝑂𝑅 = 25
⇒ 𝑂𝑅 = 5 𝑐𝑚
∴ 𝑂𝑅 = 𝑂𝑃                                          [Radii of the circle]
⇒ 𝑂𝑃 = 5 𝑐𝑚
Now, in Δ𝑂𝑃𝑀
𝑂𝑀² = 𝑂𝑃²– 𝑃𝑀²                               [Pythagoras theorem]
⇒ 𝑂𝑀² = 5² − 4² = 25 – 16 = 9
𝑂𝑀 = 9 = 3 𝑐𝑚
Hence, the distance of the other chord from the centre is 3 𝑐𝑚.

Q.4) Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ 𝐴𝐵𝐶 is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Sol.4) Given : Two equal chords AD and CE of a circle with centre O. When meet at B when produced. 
To Prove : ∠𝐴𝐵𝐶 = 1/2
(∠𝐴𝑂𝐶 – ∠𝐷𝑂𝐸)
Proof : Let ∠𝐴𝑂𝐶 = 𝑥, ∠𝐷𝑂𝐸 = 𝑦, ∠𝐴𝑂𝐷 = 𝑧
∠𝐸𝑂𝐶 = 𝑧 [Equal chords subtends equal angles at the centre]
∴ 𝑥 + 𝑦 + 2𝑧 = 36° [Angle at a point] .. (i)
𝑂𝐴 = 𝑂𝐷 ⇒ ∠𝑂𝐴𝐷 = ∠𝑂𝐷𝐴
∴ In 𝐷𝑂𝐴𝐷, we have
∠𝑂𝐴𝐷 + ∠𝑂𝐷𝐴 + 𝑧 = 180°
⇒ 2∠𝑂𝐴𝐷 = 180° – 𝑧 [ ∠𝑂𝐴𝐷 = ∠𝑂𝐵𝐴]

Q.5) Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Sol.5) Given : A rhombus ABCD whose diagonals intersect each other at O.
To prove : A circle with AB as diameter passes through O.
Proof : ∠𝐴𝑂𝐵 = 90°                      [Diagonals of a rhombus bisect each other at 90°]
⇒ Δ𝐴𝑂𝐵 is a right triangle right angled at O.
⇒ 𝐴𝐵 is the hypotenuse of A B right ΔAOB.
⇒ If we draw a circle with AB as diameter, then it will pass through O.
because angle is a semicircle is 90° and ∠𝐴𝑂𝐵 = 90° Proved.

Q.6) ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that 𝐴𝐸 = 𝐴𝐷.
Sol.6) Given : ABCD is a parallelogram.
To Prove : AE = AD.
Construction : Draw a circle which passes through ABC and intersect CD (or CD produced) at E.
Proof : For fig (i),
∠𝐴𝐸𝐷 + ∠𝐴𝐵𝐶 = 180° [Linear pair] ... (ii)
But ∠𝐴𝐶𝐷 = ∠𝐴𝐷𝐶 = ∠𝐴𝐵𝐶 + ∠𝐴𝐷𝐸
⇒ ∠𝐴𝐵𝐶 + ∠𝐴𝐷𝐸 = 180° [From (ii)] ... (iii)
From (i) and (iii)
∠𝐴𝐸𝐷 + ∠𝐴𝐵𝐶 = ∠𝐴𝐵𝐶 + ∠𝐴𝐷𝐸
⇒ ∠𝐴𝐸𝐷 = ∠𝐴𝐷𝐸
⇒ ∠𝐴𝐷 = ∠𝐴𝐸 [Sides opposite to equal angles are equal]
Similarly we can prove for Fig (ii) Proved.

Q.7) AC and BD are chords of a circle which bisect each other. Prove that
(i) AC and BD are diameters, (ii) ABCD is rectangle.
Sol.7) Given : A circle with chords AB and CD which bisect each other at O.
To Prove : (i) AC and BD are diameters
(ii) ABCD is a rectangle.
Proof : In Δ𝑂𝐴𝐵 and Δ𝑂𝐶𝐷, we have
OA = OC               [Given]
OB = OD               [Given]
∠𝐴𝑂𝐵 = ∠𝐶𝑂𝐷 [Vertically opposite angles]
⇒ Δ𝐴𝑂𝐵 ≅ ∠𝐶𝑂𝐷 [SAS congruence]
⇒ ∠𝐴𝐵𝑂 = ∠𝐶𝐷𝑂 and ∠𝐵𝐴𝑂 = ∠𝐵𝐶𝑂 [CPCT]
⇒ 𝐴𝐵 || 𝐷𝐶 ... (i)
Similarly, we can prove 𝐵𝐶 || 𝐴𝐷 ... (ii)
Hence, ABCD is a parallelogram.
But ABCD is a cyclic parallelogram.
∴ ABCD is a rectangle. [Proved in Q. 12 of Ex. 10.5]
⇒ ∠𝐴𝐵𝐶 = 90° and ∠𝐵𝐶𝐷 = 90°
⇒ AC is a diameter and BD is a diameter [Angle in a semicircle is 90°] Proved.

Q.8) Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle 𝐷𝐸𝐹 are 90°−(1/2)𝐴, 90°–(1/2)𝐵 and 90°−(1/2)𝐶 –
Sol.8) Given : ΔABC and its circumcircle. AD, BE, CF are bisectors of ∠A, ∠B, ∠C respectively.
Construction : Join DE, EF and FD.
Proof : We know that angles in the same segment are equal.

Q.9) Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Sol.9) Given : Two congruent circles which intersect at A and B. PAB is a line through A.
To Prove : BP = BQ.
Construction : Join AB.
Proof : AB is a common chord of both the circles.
But the circles are congruent
⇒ 𝑎𝑟𝑐 𝐴𝐷𝐵 = 𝑎𝑟𝑐 𝐴𝐸𝐵
⇒ ∠𝐴𝑃𝐵 = ∠𝐴𝑄𝐵 Angles subtended
⇒ 𝐵𝑃 = 𝐵𝑄                               [Sides opposite to equal angles are equal] Proved.

Q.10) In any triangle ABC, if the angle bisector of ∠𝐴 and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Sol.10) Let angle bisector of ∠𝐴 intersect circumcircle of Δ𝐴𝐵𝐶 at D.
Join DC and DB.
∠𝐵𝐶𝐷 = ∠𝐵𝐴𝐷                                      [Angles in the same segment]
⇒ ∠𝐵𝐶𝐷 = ∠𝐵𝐴𝐷 1/2
∠𝐴 [AD is bisector of ∠𝐴] ...(i)
Similarly ∠𝐷𝐵𝐶 = ∠𝐷𝐴𝐶 1/2
∠𝐴 ... (ii)
From (i) and (ii) ∠𝐷𝐵𝐶 = ∠𝐵𝐶𝐷
⇒ 𝐵𝐷 = 𝐷𝐶                                             [sides opposite to equal angles are equal]
⇒ 𝐷 lies on the perpendicular bisector of BC.
Hence, angle bisector of ∠𝐴 and perpendicular bisector of BC intersect on the circumcircle of Δ𝐴𝐵𝐶    Proved.