**Chapter 10: Circles Exercise – 10.2**

**Question: 1**

If PT is a tangent at T to a circle whose center is 0 and OP = 17 cm, OT = 8 cm. Find the length of tangent segment PT.

**Solution:**

OT = radius = 8 cm

OP = 17 cm

PT = length of tangent =?

T is point of contact. We know that at point of contact tangent and radius are perpendicular.

∴ OTP is right angled triangle ∠OTP = 90°, from Pythagoras theorem OT^{2} + PT^{2} = OP^{2}

8^{2} + PT^{2 }= 17^{2}

∴ PT = length of tangent = 15 cm.

**Question: 2**

Find the length of a tangent drawn to a circle with radius 5cm, from a point 13 cm from the center of the circle.

**Solution:**

Consider a circle with center O.

OP = radius = 5 cm.

A tangent is drawn at point P, such that line through O intersects it at Q, OB = 13cm.

Length of tangent PQ =?

A + P, we know that tangent and radius are perpendicular.

∆OPQ is right angled triangle, ∠OPQ = 90°

By Pythagoras theorem, OQ^{2} = OP^{2} + PQ^{2}

⇒ 13^{2} = 5^{2} + PQ^{2}

⇒ PQ^{2} = 169 - 25 = 144

⇒ PQ = √114

= 12 cm

Length of tangent = 12 cm

**Question: 3**

A point P is 26 cm away from 0 of circle and the length PT of the tangent drawn from P to the circle is 10 cm. Find the radius of the circle.

**Solution:**

Given OP = 26 cm

PT = length of tangent = 10 cm

radius = OT = ?

At point of contact, radius and tangent are perpendicular ∠OTP = 90°, ∆OTP is right angled triangle. By Pythagoras theorem,

OP^{2} = OT^{2} + PT^{2}

26^{2} = OT^{2} + 10^{2}

= 24 cm

OT = length of tangent = 24 cm

**Question: 4**

If from any point on the common chord of two intersecting circles, tangents be drawn to circles, prove that they are equal.

**Solution:**

Let the two circles intersect at points X and Y.

XY is the common chord.

Suppose 'A' is a point on the common chord and AM and AN be the tangents drawn A to the circle

We need to show that AM = AN.

In order to prove the above relation, following property will be used.

"Let PT be a. tangent to the circle from an external point P and a secant to the circle through P intersects the circle at points A and B, then PT^{2} = PA × PB"

Now AM is the tangent and AXY is a secant ∴ AM^{2} = AX × AY ... (i)

AN is a tangent and AXY is a secant

∴ AN^{2} = AX × AY .... (ii)

From (i) & (ii), we have AM^{2} = AN^{2}

∴ AM = AN

**Question: 5**

If the quadrilateral sides touch the circle prove that sum of pair of opposite sides is equal to the sum of other pair.

**Solution:**

Consider a quadrilateral ABCD touching circle with center O at points E, F, G and 11 as in figure.

We know that

The tangents drawn from same external points to the circle are equal in length.

1. Consider tangents from point A [AM ⊥ AE]

AH = AE ... (i)

2. From point B [EB & BF]

BF = EB ... (ii)

3. From point C [CF & GC]

FC = CG ... (iii)

4. From point D [DG & DH]

DH = DG .... (iv)

Adding (i), (ii), (iii), & (iv)

(AH + BF + FC + DH) = [(AC + CB) + (CG + DG)]

⟹ (AH + DH) + (BF + FC) = (AE + EB) + (CG + DG)

⟹ AD + BC = AB + DC [from fig.]

Sum of one pair of opposite sides is equal to other.

**Question: 6**

If AB, AC, PQ are tangents in Fig. and AB = 5 cm find the perimeter of ∆APQ.

**Solution:**

Perimeter of AAPQ, (P) = AP + AQ + PQ

= AP + AQ + (PX + QX)

We know that

The two tangents drawn from external point to the circle are equal in length from point A,

AB = AC = 5 cm

From point P, PX = PB

From point Q, QX = QC

Perimeter (P) = AP + AQ + (PB + QC)

= (AP + PB) + (AQ + QC)

= AB + AC = 5 + 5

= 10 cm.

**Question: 7**

Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at center.

**Solution:**

Consider circle with center 'O' and has two parallel tangents through A & B at ends of diameter.

Let tangents through M intersects the tangents parallel at P and Q required to prove is that LPOQ = 90°. From fig. it is clear that ABQP is a quadrilateral

∠A + ∠B = 90° + 90° = 180° [At point of contact tangent & radius are perpendicular]

∠A + ∠B + ∠P + ∠Q = 360° [Angle sum property]

∠P + ∠Q = 360° - 180° = 180° ... (i)

At P & Q ∠APO = ∠OPQ =1/2 ∠P

∠BQO = ∠PQO = 1/2 ∠Q in (i)

2∠OPQ + 2∠PQO = 180°

∠OPQ + ∠PQO = 90° ... (ii)

In ∆OPQ, ∠OPQ + ∠PQO + ∠POQ = 180° [Angle sum property]

90° + ∠POQ = 180° [from (ii)]

∠POQ = 180° – 90° = 90°

∠POQ = 90°

**Question: 8**

In Fig below, PQ is tangent at point R of the circle with center O. If ∠TRQ = 30°. Find LPRS.

**Solution:**

Given ∠TRQ = 30°.

At point R, OR ⊥ RQ.

∠ORQ = 90°

⟹ ∠TRQ + LORT = 90°

⟹ ∠ORT = 90°- 30° = 60°

ST is diameter, LSRT = 90° [ ∵ Angle in semicircle = 90°]

∠ORT + ∠SRO = 90°

∠SRO + ∠PRS = 90°

∠PRS = 90°- 30° = 60°

**Question: 9**

If PA and PB are tangents from an outside point P. such that PA = 10 cm and LAPB = 60°. Find the length of chord AB.

**Solution:**

AP = 10 cm ∠APB = 60°

Represented in the figure

We know that

A line drawn from center to point from where external tangents are drawn divides or bisects the angle made by tangents at that point ∠APO = ∠OPB =1/2 × 60° = 30°

The chord AB will be bisected perpendicularly

∴ AB = 2AM

In ∆AMP,

AM = AP sin 30°

AP/2 = 10/2 = 5cm

AP = 2 AM =10 cm ... Method (i)

In ∆AMP, ∠AMP = 90°, ∠APM = 30°

∠AMP + ∠APM + ∠MAP = 180°

90° + 30° + ∠MAP = 180°

∠MAP = 180°

In ∆PAB, ∠MAP = ∠BAP = 60°, ∠APB = 60°

We also get, ∠PBA = 60°

∴ ∆PAB is equilateral triangle

AB = AP = 10 cm ... Method (ii)