### Chapter-10: Circles

**Que 1.** Prove that the line segments joining the points of contact of two parallel tangents is a diameter of the circle.

**Ans 1. **Consider the circle with center at O

PQ & RS are two parallel tangents to it touching at A and B respectively.

Join OA and OB

Now OA perpendicular to OQ (∴ radius is perpendicular to tangent)

And OB perpendicular to RS

∴ OA ∥ OB

But OA and OB pass through O

∴ AB is straight line through center

∴ AB is a diameter

**Que 2. **O is the centre of the circle and BCD is a tangent to it at C. Prove that ∠BAC + ∠ACD = 90 0.

**Ans 2. **Given: In the above figure, O is the centre of the circle and BCD is tangent to it at C.

To prove: ∠BAC + ∠ACD = 90°

Proof: In ΔOAC

OA = OC [radii of same circle]

⇒ ∠OCA = ∠OAC [angles opposite to equal sides are equal]

⇒ ∠OCA = ∠BAC [1]

Also,

OC ⊥ BD [Tangent at any point on a circle is perpendicular to the radius through point of contact]

⇒ ∠OCD = 90°

⇒ ∠OCA + ∠ACD = 90°

⇒ ∠BAC + ∠ACD = 90° [From 1]

**Hence Proved**

**Que 3. **In the figure quadrilateral, ABCD is drawn to circumscribe a circle. Prove that AD+BC = AB + CD.

**Ans 3. **Given:- Let ABCD be the quadrilateral circumscribing the circle with centre O. The quadrilateral touches the circle at point P,Q,R and S.

To prove:- AB + CD + AD + BC

Proof:-

As we know that, length of tangents drawn from the external point are equal.

Therefore,

AP = AS…..(1)

BP = BQ…..(2)

CR = CQ…..(3)

DR = DS…..(4)

Adding equation (1),(2),(3) and (4), we get

AP + BP + CR + DR = AS = BQ + CQ + DS

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

**Hence proved.**

**Que 4. **Prove that the tangents drawn at the end-points of the diameter of a circle are parallel.

**Ans 4. **To prove: PQ ∣∣ RS

Given: A circle with centre O and diameter AB. Let PQ be the tangent at point A & Rs be the point B.

Proof: Since PQ is a tangent at point A.

OA ⊥ PQ(Tangent at any point of circle is perpendicular to the radius through point of contact).

∠OQP = 90^{o} …………(1)

OB ⊥ RS

∠OBS = 90^{o} ……………(2)

From (1) & (2)

∠OAP = ∠OBS

i.e., ∠BAP = ∠ABS

for lines PQ & RS and transversal AB

∠BAP = ∠ABS i.e., both alternate angles are equal.

So, lines are parallel.

therefore PQ || RS.

**Que 5. **Two concentric circles have centre O, OP = 4cm, OB = 5cm. AB is a chord of the outer circle and tangent to the inner circle at P. Find the length of AB.

**Ans 5. **

OP = 4 cm, OB = 5 cm

We know that the radius is perpendicular to the tangent at the point of contact.

∠OPB = 90^{o}

In right triangle OPB,

OB^{2} = OP^{2} + PB^{2}

(5)^{2} = (4)^{2} + PB^{2}

PB^{2} = 25 – 16 = 9

PB = 3 cm

We know that perpendicular from the centre to the chord bisect the chord.

AB = 2PB = 6 cm

**Que 6. **Two tangents PA and PB are drawn to a circle with centre O such that ∠APB=120^{o}. Prove that OP = 2AP

**Ans 6. **O is the center of the given circle

∠OAP = ∠OBP = 90^{o}

(Radius is perpendicular to the tangent at the point of contact)

OA = OB

(radius of the circle)

∴ △OAP is congruent to △OBP

So that,

∠OPA = ∠OPB = 120/2 = 60^{o}

In △OAP

cos∠OPA = cos60^{o }= AP/OP {cos60^{o }= 1/2}

1/2 = AP/OP

OP = 2AP Hence proved

**Que 7. **In the isosceles triangle ABC in fig. AB = AC, show that BF = FC.

**Ans 7. **ABC is an isosceles triangle (given) AB = AC (given) BE and CF are two medians (given) To prove: BE = CF In △CFB and △BEC CE = BF (Since, AC = AB = AC/2 = AB/2 = CE = BF) BC = BC (Common) ∠ECB = ∠FBC (Angle opposite to equal sides are equal) By SAS theorem: △CFB ≅ △BEC So, BE = CF (By c.p.c.t)

**Que 8. **In the fig. a circle is inscribed in a ∆ABC with sides AB = 12cm, BC = 8 cmand AC=10cm. Find the lengths of AD, BE and CF.

**Ans 8. **We know that AD = AF

BD = BE

CE = CF

Let AD = AF = x

BD = BE = y

CE = CF = z

Then x + y = 12

y + z = 8

x + z = 10

On Solving above equation we get x = 7,y = 5,z = 3

So AD = 7 , BE = 5 , CF = 3

**Que 9. **In fig. circle is inscribed in a quadrilateral ABCD in which ∠B = 90^{o}. If AD = 23cm, AB = 29cm, and DS = 5cm, find the radius ‘r’ of the circle.

**Ans 9. **In the figure. AB, BC, CD and DA are the tangents drawn to the circle at Q, P, S and R respectively.

∴ DS = DR (tangents drawn from a external point D to the circle).

but DS = 5 cm (given)

∴ DR = 5 cm

In the fig. AD = 23 cm, (given)

∴ AR = AD – DR = 23 – 5 = 18 cm

but AR = AQ

(tangents drawn from an external point A to the circle)

∴ AQ = 18 cm

If AQ = 18 cm then (given AB = 29 cm)

BQ = AB – AQ = 29 – 18 = 11 cm

In quadrilateral BQOP,

BQ = BP (tangents drawn from an external point B)

OQ = OP (radii of the same circle)

∠QBP=∠QOP=90^{o} (given)

∠OQB=∠OPB=90^{o} (angle between the radius and tangent at the point of contact.)

∴ BQOP is a square.

∴ radius of the circle, OQ = 11 cm

**Que 10. **In fig. two circles touch each other externally at C. Prove that the common tangent at C bisects the other two tangents.

**Ans 10. **We know that lengths drawn from an external point to a

circle are equal

RP = RC and RC = RQ

RP = RQ

R is the mid-point of PQ.

RP – RO

R is the mid-point of PQ,