**CLASS 10 MATHEMATICS**

### Ch 10 – CIRCLES

Ex 10.2

**In Q.1 to 3, choose the correct option and give justification.**

**1. 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**

**Answer: **Option (A)

If OP is radius which is perpendicular to the tangent PQ. Q is the external point which is joined to the centre O.

So OP PQ

From the figure,

We have got △OPQ is a right angled triangle.

It is given that

OQ = 25 cm

and PQ = 24 cm

By using Pythagoras theorem in △OPQ,

OQ^{2} = OP^{2} + PQ^{2}

(25)^{2 }= OP^{2}+ (24)^{2 }

625 = OP² + 576

OP^{2} = 625 – 576

OP^{2} = 49

OP = 7 cm

Therefore,

option A (7 cm) is the radius of the circle.

**2. In Fig. 10.11, if TP and TQ are the 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°**

**Answer: **Option (B) 70

From the question,

OP and OQ are radius of the circle which are perpendicular to tangents PT and TQ.

OP ⊥ PT and TQ ⊥ OQ

So, In Quadrilateral OPTQ

We know that the sum of the interior angles is 360.

∠OPT = ∠OQT = 90°

∠PTQ+∠POQ+∠OPT+∠OQT = 360°

putting the respective values we get,

∠PTQ +90°+110°+90° = 360°

∠PTQ = 70°

Therefore,

∠PTQ is 70° which is option B (70°).

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

**(A) 50°**

**(B) 60°**

**(C) 70°**

**(D) 80°**

**Answer: **Option A (50°).

According to the question, OA and OB are radius of the circle which are perpendicular to tangents PB and PA.

So OA ⊥ PA and OB ⊥ PB

thus,

∠OBP = ∠OAP = 90°

Now,

in the quadrilateral AOBP,

The sum of all the interior angles will be 360°

thus,

∠AOB+∠OAP+∠OBP+∠APB = 360°

Putting their values,

∠AOB + 90 + 90 + 80 = 360

∠AOB + 260° = 360°

∠AOB = 100°

Now,

The triangles △OPB and △OPA.

Here,

AP = BP (Since the tangents from a point are always

equal)

OA = OB (Which are the radii of the circle)

OP = OP (It is the common side)

Now, we can say that triangles OPB and OPA are

Similar using SSS congruency.

∴△OPB ≅ △OPA

So, ∠POB = ∠POA

∠AOB = ∠POA + ∠POB

2 (∠POA) = ∠AOB

Putting the respective values,

We get,

- ∠POA = 100°/2
- = 50°

As ∠POA is 50°

Therefore, answer is Option A (50°).

**4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.**

**Answer: **

Construction – Draw a circle centered O with diameter PQ and draws two tangents AB and CD at points P and Q respectively.

To Prove – AB || CD

Now, both radii i.e. OP and OQ are perpendicular to

the tangents AB and CD respectively.

So,

∠OPA = ∠OQC = ∠OPB = ∠OQD = 90°

So

∠APO = ∠DQO

and ∠BPO = ∠OQC (Since they are also alternate interior angles)

So,

We can say that line PQ and RS are parallel to each other.

PQ || RS. (Hence Proved).

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

**Solution:**

** **

Construction – Draw a circle with center O and a tangent AB which passes through L. Join O and L. Which will make radius OL.

Also let another center which is at point P. Join P to L. As PL also a radius.

**To prove :** O and P are coincident each other.

Since tangent at a point to a circle is perpendicular to the radius through the point.

OL AB So ∠OLB = 90 ͦ ………(1)

But if PL is also radius of circle then

∠PLB = 90 ͦ (by construction) ………..(2)

So by equation (1) and (2)

∠OLB = ∠PLB

But this is not possible that from same tangent point there is two perpendiculars.

Means O and P point are same points or we can say they coinside each other.

So we can say that –

**the perpendicular at the point of contact to the tangent to a circle passes through the center.**

**6. 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.**

**Answer:**

According to the diagram PA is the tangent of circle centered O, Join OA. And OP is the radius.

So, OP ⊥ PA

We know, OA (hypotenuse) = 5cm and

PA (perpendicular) = 4 cm

Now, △PAO is a right triangle,

OA^{2} = PA^{2 } + OP^{2}(Using Pythagoras theorem)

5^{2 }= 4^{2 }+ OP^{2 }

25 = 16 + OP^{2}

OP^{2} = 25-16

OP^{2} = 9

OP = 3 cm

So, the radius of the given circle i.e. OP is 3 cm.

**7. 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.**

**Answer:**

Draw two concentric circles with center O.

And also, draw a chord AB in the larger circle which

touches the smaller circle at a point L as shown

in the figure below.

According to the figure,

AB is tangent to the smaller circle to point L.

∴ OL ⊥ AB

Using Pythagoras theorem in triangle OPA,

OA^{2}= AL^{2}+OL^{2}

5^{2} = AL^{2}+3^{2}

AL^{2} = 25-9

AL = 4 cm

Now, AL = LB ( OL ⊥ AB, and The perpendicular from the center of the circle bisects the chord)

So, AB = 2AL

= 2×4

= 8 cm

So, the length of the chord of the larger circle

is 8 cm.

**8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC**

**Answer:**

By the figure, we can get

DR = DS …………(1) the lengths of tangents drawn from an external point to a circle are equal.

BP = BQ …………(2)

AP = AS …………(3)

CR = CQ …………(4)

Now,

adding LHS and RHS of equation (1),(2),(3),(4)

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

By rearranging

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

By simplifying,

AD+BC= CD+AB

So,

AD+BC= CD+AB

**9. In Fig. 10.13, 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°.**

**Answer: **

In the above figure, we will join OC

Then we will get-

Now, In triangles △OPA and △OCA

OP = OC (Radii of the same circle)

AO = AO (Common side)

AP = AC (length of tangents are equal from point A)

So, by sss congruency

△OPA ≅ △OCA

Similarly,

△OQB ≅ △OCB

So,

∠POA = ∠COA … (Eq i)

And,

∠QOB = ∠COB … (Eq ii)

Since,

the line POQ is a straight line,

it can be considered as a diameter of the circle.

So, ∠POA +∠COA +∠COB +∠QOB = 180°

Now, from equations (i) and (ii) we get,

2∠COA + 2∠COB = 180°

∠COA + ∠COB = 90°

∴∠AOB = 90°

**10. 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 center.**

**Answer:**

First,

draw a circle with centre O. From an external point B draw two tangents BC and BA at point A and C where OC and OB are radius of circle.

Now, The diagram is as follows:

We get quadrilateral OABC

In which ∠ OCB = 90 and ∠OAB = 90

We know that the tangent at any point of a circle is perpendicular to radius through the point of contact.

Now,

in the quadrilateral OAPB,

∴∠OCB+∠COA +∠OAB +∠ABC = 360°

(Since the sum of all interior angles of quadrilateral is 360°)

By putting the values we get,

90 + ∠COA + 90 + ∠ABC = 360°

180 + ∠COA + ∠ABC = 360°

Therefore,

∠COA + ∠ABC = 180° (Hence proved).

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

**Answer:**

As in the figure let a parallelogram ABCD which is circumscribing a circle with a center O.

Now, ABCD is a parallelogram,

So AB = CD and BC = AD.

And we know that,

DS = DP …… (1) The lengths of tangents drawn from an external point to a circle are equal.

AS = AR …… (2)

BQ = BR …… (3)

CQ = CP …… (4)

Adding equations (1), (2), (3), (4) we get,

DS + AS + BQ + CQ = DP + AR + BR + CP

By rearranging, we get,

(DS + AS) + (BQ + CQ) = (DP + CP) + (AR + BR)

Again by rearranging them we get,

AD + BC = CD + AB

here

AB = CD and BC = AD,

the above equation becomes

2AD = 2CD

∴ AD = CD

Since,

AB = BC = CD = DA,

it can be said that ABCD is a rhombus.

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

**Solution:**

From the figure:

ABC is given, And a circle is inscribed in the triangle.

CD = 6 cm , BD = 8 cm.

and,

CF = CD = 6 cm (The lengths of tangents drawn from an external point to a circle are equal.)

BE = BD = 8 cm

Now from the above figure let AF be x cm , So –

AE = AF = x

We can say that,

AB = EB+AE

= 8+x

CA = CF+FA

= 6+x

BC = DC+BD

= 6+8 = 14

We will calculate now are of triangles by herons formula

s = (AB+CA+BC)/2 (Where s = semi perimeter of triangle)

By putting the values we get,

s = (x + 8 + x + 6 + 14)/2

s = (2x + 28)/ 2

s = x + 14

Where a = BC, b = AC, c = AB

By putting the values,

= √

=

= ……… (i)

Again,

the area of △ABC

= area of (△AOC + △BOC + △COA)

= [(½×AC×OF)+(½×BC×OD)+(½×AC×OE)]

= [(½×(x+8)×4)+ (½×(x+6)×4) + ([(½×(14)×4)

= [ 2x + 16 + 2x + 12 + 28 ]

= 56+4x ………….. (ii)

Now from (i) and (ii) we get,

= = 56+4x

squaring both the sides,

48x(14+x) = (56+4x)^{2}

48x = [4(14+x)]^{2}/(14+x)

48x = 16(14+x)

48x = 224+16x

32x = 224

x = 7 cm

AB = 8+x

= 8 + 7 = 15 cm

And, CA = x+6

= 7+6 = 13 cm,

So answer is AB = 15 cm and CA = 13 cm.

**13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.**

**Solution: To prove – **

Construct a quadrilateral ABCD and circumscribed a circle with centre O which touches the circle at point P, Q, R, and S. Now, after joining the vertices of ABCD

We will get the figure:

Lets take the triangles OAP and OAS,

AP = AS (The lengths of tangents drawn from an external point to a circle are equal.)

OA = OA (common side)

OP = OS (radii of the circle)

So, by SSS congruency △OAP ≅ △OAS

So, ∠POA = ∠AOS

Means ∠2 = ∠1

Similarly,

∠3 = ∠4

∠5 = ∠6

∠8 = ∠7

by adding all these angles we get,

∠1+∠2+∠3 +∠4 +∠5+∠6+∠7+∠8 = 360°

Now by rearranging,

(∠2+∠2)+(∠3+∠3)+(∠6+∠6)+(∠7+∠7) = 360°

2∠2+2∠3+2∠6+2∠7 = 360°

2(∠2+∠3+∠6+∠7) = 360°

Thus,

∠2+∠3+∠6+∠7 = 180°

(∠2+∠3)+(∠6+∠7) = 180°

∠AOB+∠COD = 180°

Similarly,

We can prove that ∠BOC+∠AOD = 180°

Therefore,

The opposite sides of any quadrilateral which is circumscribing a given circle will subtend supplementary angles at the center of the circle.