# Question-1

**How many tangents can a circle have?**

**Solution:**

Infinitely many.

# Question-2

**Fill in the blanks:**

(i) A tangent to a circle intersects it in _______________ point(s).

(ii) A line intersecting a circle in two points is called a ________.

(iii) A circle can have ________ parallel tangents at the most.

(iv) A common point of a tangent to a circle and the circle is called _______.

(i) A tangent to a circle intersects it in _______________ point(s).

(ii) A line intersecting a circle in two points is called a ________.

(iii) A circle can have ________ parallel tangents at the most.

(iv) A common point of a tangent to a circle and the circle is called _______.

**Solution:**

(i) One

(ii) Secant

(iii) Two

(iv) Point of contact.

# Question-3

**A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :**

(A) 12 cm (B) 13 cm (C) 8.5 cm (D) cm

(A) 12 cm (B) 13 cm (C) 8.5 cm (D) cm

**Solution:**

(D) cm.

# Question-4

**Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.**

**Solution:**

AB - tangent

A'B' - secant.

# Question-5

**Choose the correct option and give justification.From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is ________.**

(A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm.

(A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm.

**Solution:**

Since the radius OT is perpendicular to QT,

In a right angled OQT,

OT^{2} = OQ^{2} – QT^{2
}= (25)^{2 }– (24)^{2
}= 625 – 576

^{ }= 49 ∴ OT = 7 cm

(A) 7 cm.

# Question-6

**Choose the correct option and give justification.**

In this figure, if PT and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to ________

(A) 60° (B) 70° (C) 80° (D) 90°.

In this figure, if PT and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to ________

(A) 60° (B) 70° (C) 80° (D) 90°.

**Solution:**

Join PQ. OP and OQ are the radius.

In OPQ, ∠POQ + ∠OPQ + ∠OQP = 180°

^{ }110°

^{ }+ 2∠OPQ = 180° (Since OP = OQ, ∠OPQ = ∠OQP)

2∠OPQ = 180° - 110°

^{ }= 70°

ÐOPQ =

= 35°

But ∠OPQ = 2∠PTQ

= 70°

(B) 70°.

# Question-7

**Choose the correct option and give justification.**

If tangents PA and PB from a point to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to ______.

(A) 50° (B) 70° (C) 80° (D) 90°.

If tangents PA and PB from a point to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to ______.

(A) 50° (B) 70° (C) 80° (D) 90°.

**Solution:**

align="center">

PA = PB (Tangents drawn from the same point to a circle are equal)

PO is common

∠PAO = ∠PBO (Tangent is perpendicular to the radius at the point of contact)

ΔAPO ~ ΔBPO (ASA similarity)

∴ ∠APO = ∠BPO = 40°

In ΔAPO, ∠POA = 180° - (90° + 40° )

= 180° - 130° = 50°

# Question-8

**Solution:**

**Given**

A circle with centre 'O'. AB is the diameter. Tangents T and Q are drawn touching the circle at A and B.

**To prove**

Tangent AT || Tangent BQ

**Proof**

In the given figure ∠TAO = 90° (angle between the tangent and the radius drawn from the point of contact is 90°).

Similarly ∠QBO = 90°

They form co-interior angles whose sum is 180°.

∴ Tangent at A and Tangent at B are parallel to each other.

# Question-9

**Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.**

**Solution:**

**Given**

PT is a tangent to a circle, A is the point of contact of the tangent with

the circle

**To prove**

The perpendicular at the point of contact to the tangent to a circle passes through the centre.

**Proof**

As we know that the tangent at any point of a circle is perpendicular

to the radius through the point of contact and also the radius passes through

the centre of the circle.

Thus it is proved that the perpendicular at the point of contact to

**the**

Tangent to a circle passes through the centre.

# Question-10

**The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.**

**Solution:**

**Given**

AP is a tangent to the circle intersect at P. AP = 4 cm and OA = 5 cm.

**To find**

OP

**Proof**

In a right angled ΔOAP, by Pythagoras theorem

Therefore the length of the tangent from a point is 3 cm.

# Question-11

**Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.**

**Solution:**

Let C

_{1}, C

_{2}be the two concentric circles of radii 5 cm and 3 cm with the centre at O

and a chord AB of the larger circle C

_{1}which touches the smaller circle C

_{2 }at P.

In a right angled OAP,

AP

^{2}= 0A

^{2}– OP

^{2 }= 25 – 9

^{ }= 16

AP = 4 cm

The length of the chord AB = AP + PB

= 4 + 4 = 8 cm.

# Question-12

**A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.**

**Solution:**

**Given**

Quad. ABCD is drawn to circumscribe a circle.

**To prove**

AB + CD = AD + BC.

**Proof**

Let P, Q, R and S be the points at which the circle touches the quadrilateral ABCD.

Using the theorem, the length of the tangents drawn form an external point to a circle are equal. We have:

AP = AS

BP = BQ

CR = CQ

DR = DS

Adding above all,

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

AB + CD = AD + BC

Hence proved.

# Question-13

**In the figure XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that**∠

**AOB = 90°.**

**Solution:**

In Δ's OPA and COA,

∠OPA = ∠OCA = 90° (Tangent Radius)

OA is common.

OP = OC (radius)

∴ ΔOPA ≡ ΔOCA

∴ ∠PAO = ∠OAC -------(1)

In Δ's OBQ and OBC,

∠ OQB = ∠ BCQ = 90° (Tangent Radius)

OB is common.

OQ = OC (radius)

∴ Δ OQB ≡ Δ OCB

∴ ∠OBQ = ∠OBC ---------(2)

But XY || X′Y′ and AB is the transversal.

∠PAC + ∠ABQ = 180° (Co interior angles)

2OAC + 2∠OBC = 180° (From 1 and 2)

∴∠OAC + ∠OBC = 90°

In Δ BOA,

∠AOB + ∠OAC + ∠OBC = 180°

∠AOB + 90° = 180°

∴∠AOB = 90°

# Question-14

**Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.**

**Solution:**

^{}

**Given**

PR and PQ are two tangents to a circle from an external point P.

**To prove**∠P + ∠QOR = 180°

^{ }

**Proof**

Using the theorem, the tangent at any point of a circle is perpendicular to the radius through the point of contact. Hence

∠OQP = 90°, ÐORP = 90°

In quad. PQOR,

∠P + ∠PRO + ∠ROQ + ∠OQP = 360°

^{ }(Sum of the angles of a quad. is 360°)

ÐP + 90° + ∠ ROQ + 90° = 360°

∠P + ∠QOR = 180°.

# Question-15

**Prove that the parallelogram circumscribing a circle is a rhombus.**

**Solution:**

**Given**

ABCD is a parallelogram which touches a circle at P, Q, R and S.

**To prove**

ABCD is a rhombus

**Proof**

AB = CD; AD = BC (opposite sides of a || gm are equal) -----------(i)

AB = AP + PB

But AP = AS and PB = BQ (Tangent from an external point are equal in length).

AB = AS + BQ

Similarly, CD = CQ + DS.

AB + CD = AS + BQ + CQ + DS

2AB = (AS +DS)+ (BQ + CQ)

= AD + BC

= 2AD [from (i)]

∴ AB = AD ---------------(ii)

From (i) and (ii), AB = CD = AD = BC.

∴ ABCD is a rhombus. (Definition of a rhombus.).

# Question-16

**A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC is divided by the point of contact D are of the lengths 8 cm and 6 cm respectively. Find the sides AB and AC.**

**Solution:**

**Given**

A triangle ABC is drawn to circumscribe a circle of radius 4cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6cm respectively.

**To Find**

The sides AB and AC.

**Proof**

Length of two tangents drawn from the same point to a circle are equal

AF = AE ............(i)

CD = CF.......... (ii)

BD = BE ..........(iii)

Let us assume AF be "x". Then from

(i) AF = AE = x

Q s = where a = BC, b = CA, c = AB

BC = BD + DC

= 8 + 6

= 14 cm

CA = CF + FA

= 6 + x

AB = AE + EB

= x + 8

s =

=

=

=

= 14 + x

Area of a triangle =

where s – semi perimeter

s – a = x + 14 – (14)

s – b = x + 14 – (x +6)

s – c = x + 14 – (x + 8)

Area of a triangle

=

=

=

= ……………(i)

Area of triangle

= 2 area of Δ COD + 2 area of Δ AOF + 2 area of Δ DOB

=

= 6 × 4 + 4 × x + 8 × 4

= 24 + 4x + 32

= 56 + 4x ………………………(ii)

Equating (i) and (ii)

= 56 + 4x

squaring on both sides we get

48x(x + 14) = (56 + 4x)

^{2}

48x(x + 14) = [4(14 + x)]

^{2}

48x(x + 14) = 16(14 + x)

^{2}

48x = 16(x + 14)

48x = 16x + 224

48x - 16x = 224

32x = 224

x = 7 cm

AC = AF + FC = 6 + 7 = 13 cm. (Q AC = AF + FC = 6 + x )

AB = AE + EB = 7 + 8 = 15 cm. (Q AB = AE + EB = x + 8 )

∴ AB = 15 cm, AC = 13 cm.