# Circles – Exercise 10.6 – (MATHEMATICS) – 9th Class

**Exercise 10.6**

**1. ****Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.**

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**Sol.**Consider triangles APB and AQB,

AP = AQ [Radii]

PB = QB [Radii]

AB is common

∴ ΔAPB ≅ ΔAQB [SSS]

∴ ∠APB = ∠AQB. [CPCT]

** 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.**

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**Sol.**Let OL is drawn perpendicular to AB and LO is produced to M meeting CD at M, then LM ⊥ CD.

[ AB ǁ‖ CD]

∴ ∠ALM = 90° ; ∠OMC = 90°

Let OL = x, OA = r. Then OM = 6 – x.

Now, in right-angled triangle OLA.

……(i)

In right-angled triangle OML,

……(ii)

From (i) and (ii), we get

Substituting x = 5 in (i), we get

Hence, radius of the circle is .

**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?**

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**Sol.**Let AB = 6 cm and CD = 8 cm are two parallel chords of a circle with centre at O.

We draw OM perpendicular to AB meeting AB at M and CD at L. Then OL is also perpendicular to CD.

and

Let OL = *x.*

In right-angled triangle OMA,

OA^{2} = AM² + OM² = (3)² + (4)² = 25 ⇒ OA = 5 cm.

As OC = OA ∴ OC = 5 cm.

In right-angled triangle OLC,

OC² = CL² + OL² ∴ (5)² = (4)² + x²

⇒ x² = 25 – 16 = 9 ∴ *x* = 3

Therefore, distance of other chord is 3 cm.

**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.**

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**Sol.**

**Construction:**Join E and A.

**Proof:**∠AOC = 2∠AEC …(i)

[Angle subtended by an arc at the centre is double the angle subtended by it on the remaining part of the circle.]

∠DOE = 2∠DAE …(ii) [Reason same as above]

Also ∠AEC = ∠ABC + ∠BAE

[Exterior angle of a triangle is equal to sum of interior opposite angles]

⇒ ∠AEC = ∠ABC + ∠DAE

⇒

**5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.**

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**Sol.**We know that diagonals of a rhombus intersect each other at right angles.

∴ ∠AOB = 90°. Also we know that angle in a semicircle is 90°. If a circle is drawn with AB as diameter, then O lies on the circle.

Hence, if a circle is drawn with any sides of the rhombus as diameter, then it passes through the point of intersection of diagonals.

**6. ABCD is a parallelogram. The circle through A, B and C intersects CD (produced if necessary) at E. Prove that AE = AD.**

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**Sol.**ABCD is a parallelogram.

∴ ∠ABC = ∠CDA …(i) [Opposite angles of a parallelogram]

Also, ABCE is a cyclic quadrilateral.

∴ ∠ABC + ∠AEC = 180° …(ii)

[Sum of opposite angles of a cyclic quadrilateral is 180°]

**In Fig. I,**

Also, ∠ADC + ∠ADE = 180° [Linear pair]

⇒ ∠ABC + ∠ADE = 180° . ..(iii) [From (i)]

From (ii) and (iii), we get

∠AEC = ∠ADE ⇒ ∠AED = ∠ADE

⇒ AD = AE

[Sides opposite to equal angles of a triangle are equal.]

**In Fig. II,**

∠AEC + ∠AED = 180° …(iv) [Linear pair]

From (i) and (ii), we get

∠CDA + ∠AEC = 180° ⇒ ∠EDA + ∠AEC = 180° …(v)

From (iv) and (v), we get

∠AED = ∠EDA

⇒ AD = AE.

[Sides opposite to equal angles of a triangle are equal]

**7. AC and BD are chords of a circle which bisect each other. Prove that**

** (i) AC and BD are diameters,**

** (ii) ABCD is a rectangle.**

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**Sol.**(i) Consider triangles AOD and BOC.

AO = OC [Given]

BO = OD [Given]

∠AOD = ∠BOC [Vertically opposite angles]

⇒ ΔAOD ≅ ΔCOB [SAS]

⇒ ∠DAO = ∠BCO

Also, ∠ADO = ∠BCO

[Angles in the same segment of a circle]

From (i) and (ii), we have

∠DAO = ∠ADO

⇒ OA = OD

Hence, OA = OB = OC = OD.

⇒ 0 is equidistant from A, B, C, D.

⇒ 0 is centre of the circle.

Hence, AC and BD are diameters.

(ii) As AC and BD are diameters. Therefore, ∠BAD,∠ABC, ∠BCD, ∠CDA are angles in a semicircle.

Therefore, each of them is 90°.

Hence, ABCD is a rectangle.

**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 and**

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**Sol.**∠ADE = ∠ABE …(i)

[Angles in the same segment of a circle]

∠ADF = ∠ACF …(ii) [Reason same as above]

Now ∠EDF = ∠ADE + ∠ADF

⇒ ∠EDF = ∠ABE + ∠ACF

…(iii) [From (i), (ii)]

As BE and CF are bisectors of ∠ABC and ∠ACB.

and ….(iv)

From (iii) and (iv), we get

…..(v)

In triangle ABC,

∠ABC + ∠ACB + ∠BAC =180°

⇒ ∠ABC + ∠ACB = 180° – ∠BAC.

Substituting in (v), we get

Similarly, we can show that

**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.
**

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**Sol.**As AB is common chord.

∴ arc AXB = arc AYB

⇒ ∠AQB = ∠APB

[Equal areas of congruent circles subtend equal angles on the remaining part of the circles.]

Now in triangle BPQ,

∠AQB = ∠APB [Proved above]

⇒ ∠PQB = ∠QPB

⇒ PB = QB.

[Sides opposite to equal angles are equal.]

**10. In any triangle ABC, if the angle bisector of LA and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.**

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**Sol.**Let 0 be the centre of the circumcircle of triangle ABC.

∴ ∠BOC = 2∠BAC

[Angle subtended by an arc at the centre is doubled the angle subtended by it at remaining part of the circle.]

Consider ΔOLB and ΔOLC.

OB = OC [Radii]

OC is common.

BL = LC [OL is perpendicular bisector]

∴ ΔOLB ≅ ΔOLC [SSS]

∴ ∠BOL = ∠COL = ∠CAB …(i)

Let perpendicular bisector of side BC and angle bisector ∠A meet a the point P.

From equation (i), we get

∠COL = ∠CAB

⇒ ∠COP = 2∠CAP [∠CAP = ∠BAP]

This proves that points C, A and P are concyclic points as ∠COP is subtended by arc CP at the centre O and ∠CAP is subtended by it at the remaining part of the circle.

Again, from equation (i), we get

∠BOL = ∠BAC

i.e., ∠BOP = 2∠BAP

This, also proves that points B, A and P are concyclic. Hence, P lies on the circumcircle of the ΔABC.