# Circle

## Exercise 10.6

### Part 1

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

Answer: The following figure shows two circles with centres O and O' and the circles are intersecting at points P and Q.

To Prove: ∠OPO' = ∠OQO'

In ΔOPO' and ΔOQO'

OP = OQ (Radii of same circle)

O'P = O'Q (Radii of same circle)

OO' = OO' (common side)

So, ΔOPO' ≅ ΔOQO' (SSS Theorem)

So, ∠OPO' = ∠OQO' Proved

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.

Answer: The following figure shows a circle with center O, in which AB and CD are parallel chords. AB = 5 cm, CD = 11 cm and perpendicular distance between AB and CD = 6 cm

Let us draw PQ the perpendicular distance between AB and CD so that this line passes through centre O.

Now, join A and C to O

In Δ APO:

AP = (AB)/2=2.5 cm (Line from center is perpendicular bisector of chord)

OA=r

OP=6-x (If OQ=x)

Applying Pythagoras theorem, we get

OA2 - OP2 = AP2

Or, r^2-(6-x)^2=2.5^2

Or, r^2-(36+x^2-12x)=6.25

Or, r^2-36-x^2+12x=6.25

Or, r^2=6.25+36+x^2-12x………………(1)

In ΔCQO:

CQ=(CD)/2=5.5 cm (Line from center is perpendicular bisector of chord)

OQ =x

Applying Pythagoras theorem, we get

OC2 = OQ2 + CQ2

Or, r^2=x^2+5.5^2

Or, r^2=x^2+30.25 ……………..(2)

From equations (1) and (2)

x^2+30.25=x^2-12x+42.25

Or, -12x+42.25=30.25

Or, -12x=30.25-42.25=-12

Or, x=1

Substituting the value of x in equation (1) or (2), we can calculate the value of r as follows:

r^2=1^2+30.25=31.25

Or, r=sqrt(31.25)=5.57 (approx)

Question 3: The lengths of two prallel 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?

Answer: The question does not specify whether the chords are on the same side of center or on opposite sides. We will solve by assuming both conditions.

This figure shows a circle with centre O. Two parallel chords AB and CD are on the same side of center and AB is 4 cm from the centre.

In ΔAPO

AO2 = AP2 + PQ2

Or, r^2=3^2+4^2

(AP = 6/2=3 cm because PO is perpendicular bisector of AB)

Or, r^2=9+16=25

Or, r=5 cm

In ΔCQO

OQ2 = CO2 - CQ2

Or, OQ^2=r^2-4^2

(CQ = 8/2=4 because OQ is perpendicualr bisector of CD)

Or, OQ^2=5^2-4^2=25016-9

Or, OQ = 3 cm

Let us now assume that the chords are on opposite sides of the centre, as shown in this figure.

In ΔAPO:

AP = 3 cm and OP = 4 cm

AO2 = 42 + 32 = 16 + 9 = 25

Or, AO = 5 cm

In ΔCQO:

CQ = 4 cm and CO = 5 cm (radius)

OQ2 = CO2 - CQ2

= 52 - 42 = 25 – 16 = 9

Or, OQ = 3 cm

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.

Answer: This figure shows ∠ABC and its sides are extended respectively to D and E so that AD and CE are equal chords of the circle.

To Prove: ∠ABC = 1/2 (∠DOE - ∠AOC)

Let us join A to E

Proof: In ΔBAE

∠DAE = ∠ABC = ∠AEC ………….(1)

(External angle of triangle is equal to sum of angles of opposite vertices)

Chord DE subtends ∠DOE at centre and ∠DAE in remaining part of the circle.

So, ∠DAE = 1/2 ∠DOE ……..(2)

Similarly, ∠AEC = 1/2 ∠AOC ………….(3)

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

1/2∠DO\E=∠AB\C+1/2∠AO\C

Or, ∠AB\C=1/2∠DO\E-1/2∠AO\C

Or, ∠AB\C=1/2(∠DO\E-∠AO\C) Proved

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

Answer: This figure shows a rhombus ABCD in which diagonals AC and BD intersect at O.

Let us draw a circle by taking AB as diameter. Let us draw PQ||DA and RS||AB. Both PQ and RS are passing through O. Now, P, Q, R and S are midpoints of DC, AB, AD and BC respectively.

Since Q is the midpoint o AB

Hence, AQ = QB ………(1)

Since AD = BC (sides of rhombus are equal)

So, 1/2AD=1/2BC

Or, RA = OQ ………(2)

So, AB = AD (sides of rhombus)

Or, 1/2AB=1/2AD

Or, AQ = AR ………….(2)

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

AQ = QB = OQ

This means that a circle drawn with Q as centre will pass through A, B and O

So, it is proved that the circle passes through the intersection O of the diagonals from rhombus ABCD.