## Chapter 11 Circles Ex 11.6

**Question 1.** **Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.** **Solution:** **Given:** Two circles with centres O and O’ which intersect each other at C and D.

**To prove:** ∠OCO’ = ∠ODO’ **Construction:** Join OC, OD, O’C and O’D **Proof:** In ∆ OCO’and ∆ODO’, we have

OC = OD (Radii of the same circle)

O’C = O’D (Radii of the same circle)

OO’ = OO’ (Common)

∴ By SSS criterion, we get

∆ OCO’ ≅ ∆ ODO’

Hence, ∠OCO’ = ∠ODO’ (By CPCT)

**Question 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.** **Solution:**

Let O be the centre of the given circle and let its radius be cm.

Draw ON ⊥ AB and OM⊥ CD since, ON ⊥ AB, OM ⊥ CD and AB || CD, therefore points N, O, M are collinear.

**Question 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 ?** **Solution:**

Let PQ and RS be two parallel chords of a circle with centre O such that PQ = 6 cm and RS = 8 cm.

Let a be the radius of circle.

**Question 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.** **Solution:**

Since, an exterior angle of a triangle is equal to the sum of the interior opposite angles.

**Question 5.** **Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.** **Solution:** **Given:** PQRS is a rhombus. PR and SQ are its two diagonals which bisect each other at right angles. **To prove:** A circle drawn on PQ as diameter will pass through O.

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

Since, ABCE is a cyclic quadrilateral, therefore

∠AED+ ∠ABC= 180°

(∵ Sum of opposite angle of a cyclic quadrilateral is 180°) .. .(i)

∵ ∠ADE + ∠ADC = 180° (EDC is a straight line)

So, ∠ADE + ∠ABC = 180°

(∵ ∠ADC = ∠ABC opposite angle of a || gm).. .(ii)

From Eqs. (i) and (ii), we get

∠AED + ∠ABC = ∠ADE + ∠ABC

⇒ ∠AED = ∠ADE

∴ In ∆AED We have

∠AED = ∠ADE

So, AD = AE

(∵ Sides opposite to equal angles of a triangle are equal)

**Question 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.** **Solution:**

(i) Let BD and AC be two chords of a circle bisect at P.

In ∆APB and ∆CPD, we get

PA = PC ( ∵ P is the mid-point of AC)

∠APB = ∠CPD (Vertically opposite angles)

and PB = PD (∵ P is the mid-point of BD)

∴ By SAS criterion

∆CPD ≅ ∆APB

∴ CD= AB (By CPCT) …(i)

∴ BD divides the circle into two equal parts. So, BD is a diameter.

Similarly, AC is a diameter.

(ii) Now, BD and AC bisect each other.

So, ABCD is a parallelogram.

Also, AC = BD

∴ ABCD is a rectangle.

**Question 8.**

**Solution:**

∵ ∠EDF = ∠EDA + ∠ADF

∵ ∠EDA and ∠EBA are the angles in the same segment of the circle.

∴ ∠EDA = ∠EBA

and similarly ∠ADF and ∠FCA are the angles in the same segment and hence

**Question 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.** **Solution:**

Let O’ and O be the centres of two congruent circles.

Since, AB is a common chord of these circles.

∴ ∠BPA = ∠BQA

(∵ Angle subtended by equal chords are equal)

⇒ BP = BQ

**Question 10.** **In any ∆ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC.** **Solution:**

(i) Let bisector of ∠A meet the circumcircle of ∆ABC at M.

Join BM and CM.

∴ ∠MBC = ∠MAC (Angles in same segment)

and ∠BCM = ∠BAM (Angles in same segment)

But ∠BAM = ∠CAM (∵ AM is bisector of ∠A)…. .(i)

∴ ∠MBC = ∠BCM

So, MB = MC (Sides opposite to equal angles are equal)

So, M must lie on the perpendicular bisector of BC

(ii) Let M be a point on the perpendicular bisector of BC which lies on circumcircle of ∆ ABC.

Join AM.

Since, M lies on perpendicular bisector of BC.

∴ BM = CM

∠MBC = ∠MCB

But ∠MBC = ∠MAC (Angles in same segment)

and ∠MCB = ∠BAM (Angles in same segment)

So, from Eq. (i),

∠BAM = ∠CAM

AM is the bisector of A.

Hence, bisector of ∠A and perpendicular bisector of BC at M which lies on circumcircle of ∆ABC.