**Circles**

**Question 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.**

**Solution:**

We have two intersecting circles with centres at O and O’ respectively. Let PQ be the common chord.

∵ In two intersecting circles, the line joining their centres is perpendicular bisector of the common chord.

∴∠OLP = ∠OLQ = 90° and PL = LQ

Now, in right ∆OLP, we have

PL^{2} + OL^{2} = ^{2}

⇒ PL^{2} + (4 – x)^{2} = 5^{2}

⇒ PL^{2} = 5^{2} – (4 – x)^{2}

⇒ PL^{2} = 25 -16 – x^{2} + 8x

⇒ PL^{2} = 9 – x^{2} + 8x …(i)

Again, in right ∆O’LP,

PL^{2} = PO^{‘2} – LO^{‘2}

= 3^{2} – x^{2} = 9 – x^{2} …(ii)

From (i) and (ii), we have

9 – x^{2} + 8x = 9 – x^{2}

⇒ 8x = 0

⇒ x = 0

⇒ L and O’ coincide.

∴ PQ is a diameter of the smaller circle.

⇒ PL = 3 cm

But PL = LQ

∴ LQ = 3 cm

∴ PQ = PL + LQ = 3cm + 3cm = 6cm

Thus, the required length of the common chord = 6 cm.

**Question 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.**

**Solution:**

Given: A circle with centre O and equal chords AB and CD intersect at E.

To Prove: AE = DE and CE = BE

Construction : Draw OM ⊥ AB and ON ⊥ CD.

Join OE.

Proof: Since AB = CD [Given]

∴ OM = ON [Equal chords are equidistant from the centre]

Now, in ∆OME and ∆ONE, we have

∠OME = ∠ONE [Each equal to 90°]

OM = ON [Proved above]

OE = OE [Common hypotenuse]

∴ ∆OME ≅ ∆ONE [By RHS congruence criteria]

⇒ ME = NE [C.P.C.T.]

Adding AM on both sides, we get

⇒ AM + ME = AM + NE

⇒ AE = DN + NE = DE

∵ AB = CD ⇒ 12AB = 12DC

⇒ AM = DN

⇒ AE = DE …(i)

Now, AB – AE = CD – DE

⇒ BE = CE …….(ii)

From (i) and (ii), we have

AE = DE and CE = BE

**Question 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.**

**Solution:**

Given: A circle with centre O and equal chords AB and CD are intersecting at E.

To Prove : ∠OEA = ∠OED

Construction: Draw OM ⊥ AB and ON ⊥ CD.

Join OE.

Proof: In ∆OME and ∆ONE,

OM = ON

[Equal chords are equidistant from the centre]

OE = OE [Common hypotenuse]

∠OME = ∠ONE [Each equal to 90°]

∴ ∆OME ≅ ∆ONE [By RHS congruence criteria]

⇒ ∠OEM = ∠OEN [C.P.C.T.]

⇒ ∠OEA = ∠OED

**Question 4.If a line intersects two concentric circles (circles with the same centre) with centre 0 at A, B, C and D, prove that AB = CD (see figure).**

**Solution:**Given : Two circles with the common centre O.

A line D intersects the outer circle at A and D and the inner circle at B and C.

To Prove : AB = CD.

Construction:

Draw OM ⊥ l.

Proof: For the outer circle,

OM ⊥ l [By construction]

∴ AM = MD …(i)

[Perpendicular from the centre to the chord bisects the chord]

For the inner circle,

OM ⊥ l [By construction]

∴ BM = MC …(ii)

[Perpendicular from the centre to the chord bisects the chord]

Subtracting (ii) from (i), we have

AM – BM = MD – MC

⇒ AB = CD

**Question 5.Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m 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?**

**Solution:**

Let the three girls Reshma, Salma and Mandip be positioned at R, S and M respectively on the circle with centre O and radius 5 m such that

RS = SM = 6 m [Given]

Equal chords of a circle subtend equal angles at the centre.

∴ ∠1 = ∠2

In ∆POR and ∆POM, we have

OP = OP [Common]

OR = OM [Radii of the same circle]

∠1 = ∠2 [Proved above]

∴ ∆POR ≅ ∆POM [By SAS congruence criteria]

∴ PR = PM and

∠OPR = ∠OPM [C.P.C.T.]

∵∠OPR + ∠OPM = 180° [Linear pair]

∴∠OPR = ∠OPM = 90°

⇒ OP ⊥ RM

Now, in ∆RSP and ∆MSP, we have

RS = MS [Each 6 cm]

SP = SP [Common]

PR = PM [Proved above]

∴ ∆RSP ≅ ∆MSP [By SSS congruence criteria]

⇒ ∠RPS = ∠MPS [C.P.C.T.]

But ∠RPS + ∠MPS = 180° [Linear pair]

⇒ ∠RPS = ∠MPS = 90°

SP passes through O.

Let OP = x m

∴ SP = (5 – x)m

Now, in right ∆OPR, we have

x^{2} + RP^{2} = 5^{2}

RP^{2} = 5^{2} – x^{2}

In right ∆SPR, we have

(5 – x)^{2} + RP^{2} = 6^{2}

⇒ RP^{2} = 6^{2} – (5 – x)^{2} ……..(ii)

From (i) and (ii), we get

⇒ 5^{2} – x^{2} = 6^{2} – (5 – x)^{2}

⇒ 25 – x^{2} = 36 – [25 – 10x + x^{2}]

⇒ – 10x + 14 = 0

⇒ 10x = 14 ⇒ x = 1410 = 1.4

Now, RP^{2} = 5^{2} – x^{2}

⇒ RP^{2} = 25 – (1.4)^{2}

⇒ RP^{2} = 25 – 1.96 = 23.04

∴ RP = 23.04−−−−√= 4.8

∴ RM = 2RP = 2 x 4.8 = 9.6

Thus, distance between Reshma and Mandip is 9.6 m.

**Question 6.A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting 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.**

**Solution:**

Let Ankur, Syed and David are sitting at A, S and D respectively in the circular park with centre O such that AS = SD = DA

i. e., ∆ASD is an equilateral triangle.

Let the length of each side of the equilateral triangle be 2x.

Draw AM ⊥ SD.

Since ∆ASD is an equilateral triangle.

∴ AM passes through O.

⇒ SM = 12 SD = 12 (2x)

⇒ SM = x

Now, in right ∆ASM, we have

AM^{2} + SM^{2} = AS^{2} [Using Pythagoras theorem]

⇒ AM^{2}= AS^{2} – SM^{2}= (2x)^{2} – x^{2}

= 4x^{2} – x^{2} = 3x^{2}

⇒ AM = 3x−−√m

Now, OM = AM – OA= (3x−−√ – 20)m

Again, in right ∆OSM, we have

OS^{2} = SM^{2} + OM^{2} [using Pythagoras theorem]

20^{2} = x^{2} + (3x−−√ – 20)^{2}

⇒ 400 = x^{2} + 3x^{2} – 403x−−√ + 400

⇒ 4x^{2} = 40 3x−−√

⇒ x = 10√3 m

Now, SD = 2x = 2 x 10√3 m = 20√3 m

Thus, the length of the string of each phone = 20√3 m