# NCERT Solutions for Class 10 Maths Chapter 10 - Circles

**Important Points**

**1. Circle :** A circle is a collection of all points in a plane which are at a constant distance from a fixed point. Its represent C(O, r).

Here, O = centre of circle, r = radius of circle

**2. Some Basic Definitions**

**(i) Chord :** A line segment joining any two points on a circle is called a chord of the circle.

**(ii) Diameter :** The maximum length of the chord is called the diameter of a circle.

**(iii) Secant :** A line intersect the circle in two points is calld secant of the circle.

**(iv) Tangent :** A line touches the circle at only one point is called tangent to the circle.

**(v) Cyclic quadrilateral :** If all vertices of a quadrilateral lie on a circle is called cyclic quadrilateral. Its sum of opposite angle are 180°.

**(vi) Alternate segment :** The segment opposite to the angle formed by the chord of a circle with the tangent at a point is called the alternate segment for that angle.

**(vii) Concentric circles :** Circles having the same centre.

**3. Important Theorems**

(i) The tangent at any point of a circle is perpendicular to the radius through the point of contact.

(ii) The lenghts of tangents drawn from an external point to a circle are equal.

(iii) The perpendicular from the centre of a circle to a chord bisects the chord in two equal parts and it is vice-versa.

(iv) Chords are equidistant from the centre of a circle if and only if they are equal in length.

(v) An inscribed angle subtended by a diameter (angle of semi-circle) is a right angle.

(vi) If a central angle and an inscribed angle of a circle are subtended by the same chord and on the same side of the chord, then the central angle is twice the inscribed angle.

(vii) If two angles are inscribed on the same chord and on the same side of the chord, then they are equal.

(viii) Let PT be a tangent line to the circle at point T. From external point P, draw a line which intersect the circle at two points A and B.

The, PA × PB = PT^{2}

(ix) Then opposite sides of a quadrilateral circumscribing a circle subtend suplementary angles at the centre of the circle.

**Exercise 10.1**

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

**Sol.** For every point of a circle, we can draw a tangent. Therefore, infinite tangents can be drawn.

**2. Fill in the blanks.**

**(i) A tangent to a circle touches 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) The common point of a tangent to a circle and the circle is called ............. .**

**Sol.** **(i)** One. A tangent line touch the circle only at one point.

**(ii)** Secant. Any line intersecting the circle at two points is called a secant.

**(iii)** Two. Maximum a circle can have two parallel tangents which can be drawn to the opposite side of the centre.

**(iv)** Point of contact.

**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

$$\text{(d)}\space\sqrt{119}\space\text{cm}$$

$$\textbf{Sol.\space}\text{(d)}\space\sqrt{119}\space\text{cm}$$

**Explanation:**

**Sol.** Given,

Radius OP = 5 cm

OQ = 12 cm

PQ = ?

∠OPQ = 90° (∵ The tangent at any point of a circle is perpendicular to the radius through the point of contact)

In ΔOPQ by Pythagoras theorem.

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

⇒ 12^{2} = 5^{2} + PQ^{2}

⇒ 144 = 25 + PQ^{2}

⇒ PQ^{2} = 144 – 25

⇒ PQ^{2} = 119

$$\Rarr\space\text{PQ}=\sqrt{119}\space\text{cm}$$

**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.**

**Sol.** Firstly draw a circle with centre O and draw a line l. Now, we draw a two parallel lines to l, such that one line m is tangent to the circle and another n is secant to the circle.

**Exercise 10.2**

**• 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**

**Sol.** (a) 7 cm

**Explanation:**

Given that, the length of the tangent OQ = 24 cm and QO = 25 cm. Join OP.

We know, radius OP is perpendicular to the tangent PQ.

In right ΔOPQ

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

(By Pythagoras theorem)

⇒ 25^{2} = OP^{2} + 24^{2}

⇒ OP^{2} = 625 – 576 = 49

⇒ OP = 7 cm

**2. In figure, 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°**

**Sol.** (b) 70°

**Explanation:**

Given,

∠POQ = 110°

We know that

∠OPT = ∠OQT = 90° (OP ⊥ PT, OQ ⊥ QT)

(∵ The tangent at any point of a circle is perpendicular to the radius through the point of contact)

Now, in quadrilateral POQT

∠POQ + ∠OQT + ∠QTP + ∠TPO = 360°

110° + 90° + ∠QTP + 90° = 360°

∠QTP = 360° – 110° – 90° – 90°

(∵ ∠QTP = ∠PTQ)

∠PTQ = 70°

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

**(a) 50°**

**(b) 60°**

**(c) 70°**

**(d) 80°**

**Sol.** (a) 50°

**Explanation:**

Given, two tangents PA and PB

∠APB = 80°

To find : ∠POA

**Construction :** Join OP

**Proof :** Since PA and PB are tangent.

OA ⊥ PA, OB ⊥ PB

∠OAP = 90°, ∠OBP = 90°

In ΔOAP and ΔOBP

OA = OB (Both radius)

OP = OP (Common)

∠OAP = ∠OBP = 90°

∴ ΔOAP ≅ ΔOBP (SSA congruency)

So, ∠OPA = ∠OPB

$$\angle\text{OPA}=\frac{1}{2}\angle\text{APB}\\\angle\text{OPA}=\frac{1}{2}×80\degree=40\degree$$

(Given ∠APB = 80°)

In ΔOPA

Sum of intrenal angles of Δ is 180°.

∴ ∠OPA + ∠OAP + ∠POA = 180°

40° + 90° + ∠POA = 180°

∠POA = 180° – 130°

∠POA = 50°

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

**Sol.** Let AB be a diameter of a given circle and let LM and PQ be the tangent lines drawn to the circle at points A and B, respectively. Since, the tangent at a point to a circle is perpendicular to the radius through the point.

∴ AB ⊥ PQ and AB ⊥ LM

⇒ ∠PAB = 90°

and ∠ABM = 90°

⇒ ∠PAB = ∠ABM

⇒ PQ || LM **Hence proved.**

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

**Sol.** Let be suppose that the perpendicular to AB at P does not passes through centre O, let it pass through other point O′. Join OP and O′P

∠O′PB = 90° ...(i)

O is the centre of the circle and P is the point of contact line joining the centre and the point of contact to the tangent of the circle is perpendicular to each other.

∴ ∠OPB = 90° ...(ii)

By (i) and (ii)

∠O′PB = OPB

From the figure

∠O′PB < OPB

Therefore, ∠OP′B = ∠OPB is not possible. It is only possible, when the line O′P coincides with OP.

Therefore, the perpendicular to AB through P passes through centre O.

**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.**

**Sol.** Given, OP = 5 cm and AM = 4 cm

∵ OM ⊥ AM

(∵ Radius is perpendicular to AM)

In right ΔOMA,

OP^{2} = OM^{2} + MA^{2}

(By Pythagoras theorem)

⇒ 5^{2} = OM^{2} + 4^{2}

⇒ OM^{2} = 25 – 16 = 9

⇒ OM = 3 cm

Hence, radius of the circle 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.**

**Sol.** Let O be the centre of the concentric circles. Let AB be a chord of the larger circle touching the smaller circle at P.

i.e., AD = BD

In right ΔOBD,

OB^{2} = OD^{2} + DB^{2} (∠ODB = 90°)

(By Pythagoras theorem)

[Given OB = 5 cm

OD = 3 cm]

⇒ 5^{2} = 3^{2} + DB^{2}

⇒ DB^{2} = 25 – 9 = 16

⇒ DB = 4 cm

∴ Length of chord = AB = 2 AD = 2 × 4= 8 cm

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

**Sol.** Using theorem, the lenghts of tangents drawn from an external point to a circle are equal.

Suppose, A is an external point, then

AP = AS ...(i)

Suppose, B is an external point, then

BP = BQ ...(ii)

Suppose, C is an external point, then

CQ = RC ...(iii)

Suppose, D is an external point, then

SD = RD ...(iv)

On adding Eqs. (i), (ii), (iii) and (iv), we get

(AP + BP) + (RC + RD) = (AS + BQ) + (CQ + SD)

⇒ AB + CD = (AS +SD) + (BQ + DQ)

⇒ AB + CD = AD + BC **Hence proved.**

**9. In 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°.**

**Sol.** Since, tangents drawn from an external point to a circle are equal.

∴ AP = AC

Thus, in ΔAPO and ΔACO,

AP = AC

AO = AO (Common)

OP = OC (Radius of circle)

ΔQPO ≅ ΔACO (SSS congruence)

⇒ ∠PAO = ∠OAC

⇒ ∠PAC = 2∠CAO

Similarly, we can prove that

∠CBO = ∠OBQ

⇒ ∠CBQ = 2∠CBO

Since, XY || X′Y′

∴ ∠PAC + ∠QBC = 180°

(Sum of interior angles on the same side of transversal is 180°)

∴ 2∠CAO + 2∠CBO = 180° ...(i)

⇒ ∠CAO + ∠CBO = 90°

In ΔAOB,

⇒ ∠CAO + ∠CBO + ∠AOB = 180°

⇒ ∠CAO + ∠CBO = 180° – ∠AOB

...(ii)

∴ From equations (i) and (ii), we get

180° – ∠AOB = 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 centre.**

**Sol.** Let PQ and PR be two tangents drawn from an external point P to a circle with centre O.

We have to prove that,

∠QOR = 180° – ∠QPR

or ∠QOR + ∠QPR = 180°

In right ΔOQP and ΔORP,

PQ = PR

(Length of tangents are equal drawn from an external point).

OQ = OR (Radius of circle)

OP = OP (Common)

ΔOQP = DORP (SSS congruence)

⇒ ∠QPO = ∠RPO

and ∠POQ = ∠POR

⇒ ∠QPR = 2∠OPQ

and ∠QOR = 2∠POQ ...(i)

In ΔOPQ,

∠QPO + ∠QOP = 90°

⇒ ∠QOP = 90° – ∠QPO

⇒ 2∠QOP = 180° – 2∠QPO

(Multiplying both sides by 2)

⇒ ∠QOR = 180º – ∠QPR

[From Eq . (i)]

⇒ ∠QOR + ∠QPR = 180° **Hence proved.**

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

**Sol.** Let ABCD be a parallelogram circumscribing a circle. Using the property that the tangents to a circle from an exterior point are equal in length.

∴ AM = AP ...(i)

and BM = BN ...(ii)

CO = CN ...(iii)

and DO = DP ...(iv)

By (i) + (ii) + (iii) + (iv)

(AM + BM) + (CO + DO) = AP + BN + CN + DP

$$\begin{Bmatrix} \because \text{AB = AM + MB}\\\text{BC = BN + NC}\\\text{CD=CO+OD}\\\text{AD=AP+PD}\end{Bmatrix}$$

⇒ AB + CD = (AP + PD) + (BN + NC) = AD + BC

2AB = 2BC

(∵ ABCD is a parallelogram, so opposite sides are equal)

Then, AB = CD, BC = AD

⇒ AB = BC

∴ AB = BC = CD = DA

Hence, ABCD is a rhombus.

**Hence proved.**

**12. A ΔABC 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 figure). Find the sides AB and AC.**

**Sol.** Given, CD = 6 cm, BD = 8 cm and radius = 4 cm

Join OC, OA and OB.

By using the property, tagnets drawn from external point equal in length.

∴ CD = CF = 6 cm

and BD = BE = 8 cm

Let AF = AE = x cm

In ΔOCB,

$$\text{Area of triangle, A}_1=\frac{1}{2}\text{Base×Height}\\=\frac{1}{2}×\text{CB}×\text{OD}\\=\frac{1}{2}×14×4$$

= 28 cm^{2}

In ΔOCA,

$$\text{Area of triangle, A}_2=\frac{1}{2}×\text{AC}×\text{OF}\\=\frac{1}{2}(6+x)×4$$

= 12 + 2x

In ΔOBA,

$$\text{Area of triangle, A}_3=\frac{1}{2}×\text{AB}×\text{OF}\\=\frac{1}{2}(8+x)×4$$

= 16 + 2x

Now, Perimeter of triangle,

$$Δ\text{ABC =}\frac{1}{2}(\text{AB+BC+CA})\\\text{S}=\frac{1}{2}(x+8+14+6+x)\\\text{S}=\frac{1}{2}×2(14+x)$$

S = 14 + x

Now, area of ΔABC = A_{1} + A_{2} + A_{3}

= 28 + (12 + 2x) + (16 + 2x)

= 56 + 4x ...(i)

Using Heron’s formula,

Area of ΔABC

$$=\sqrt{s(s-a)(s-b)(s-c)}\\\sqrt{(14+x)(14+x-14)(14+x-x-6)(14+x-x-8)}\\=\sqrt{(14+x)(x)(8)(6)}\\=\sqrt{(14+x)×48}$$

...(ii)

∴ From equations (i) and (ii), we get

$$\sqrt{(14+x)×48}\\=56+4x$$

On squaring both sides,

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

$$\Rarr\space\frac{48x}{16}=\frac{(14+x)^{2}}{14+x}$$

⇒ 3x = 14 + x

⇒ 2x = 14 ⇒ x = 7

∴ Length AC = 6 + x = 6 + 7 = 13 cm

Length of AB = 8 + x = 8 + 7 = 15 cm

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

**Sol.** Let ABCD be a quadrilateral circumscribing a circle with centre O, A circle touches the sides of a quadrilateral at points E, F, G and H.

**To prove:** ∠AOB + ∠COD = 180°

and ∠AOD + ∠BOC = 180°

**Construction:** Join OH, OE, OF and OG.

**Proof :** Using the property, two tangents drawn from an external point to a circle subtend equal angles at the centre.

$$\therefore\begin{drcases}\angle1=\angle2\\\angle3=\angle4\\\angle5=\angle6\\\angle7=\angle8\end{drcases}\space\text{...(i)}$$

We know the sum of all angles subtended at a point O is 360°.

∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°

⇒ 2(∠2 + ∠3 + ∠6 + ∠7) = 360° ...(ii)

⇒ (∠2 + ∠3) + (∠6 + ∠7) = 180°

⇒ ∠AOB + ∠COD = 180°

Again from equation (ii)

∠2 + ∠3 + ∠6 + ∠7 = 180°

From equation (i)

(∠1 + ∠8) + (∠4 + ∠5) = 180°

and ∠AOD + ∠BOC = 180°

**Hence proved.**

**Selected NCERT Exemplar Problems**

**Exercise 10.1**

• Choose the correct answer from the given four options.

**1. If radii of two concentric circles are 4 cm and 5 cm, then length of each chord of one circle which is tangent to the other circle is**

**(a) 3 cm**

**(b) 6 cm**

**(c) 9 cm**

**(d) 1 cm**

**Sol.** (b) 6 cm

**Explanation:**

Let O be the centre of two concentric circles C_{1} and C_{2}, whose radii are r_{1} = 4 cm and r_{2 }= 5 cm. Now, we draw a chord AC which touches a circle C_{1 }at B.

Also, join OB, which is perpendicular to AC

∴ In right ΔOAB,

OA^{2} = AB^{2} + BO^{2}

(By Pythagoras theorem)

⇒ 5^{2} = AB^{2} + 4^{2}

⇒ AB = 25 – 16 = 9

⇒ AB = 3 cm

∴ Length of chord AC = 2 AB = 2 × 3 = 6 cm

**2. In figure, if ∠AOB = 125°, then ∠COD is equal to**

**(a) 62.5°**

**(b) 45°**

**(c) 35°**

**(d) 55°**

**Sol.** (d) 55°

**Explanation:**

The opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

i.e., ∠DOC = 180° – ∠AOB

= 180° – 125° – 55°

**3. In figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to**

**(a) 45°**

**(b) 60°**

**(c) 50°**

**(d) 55°**

**Sol.** (c) 50°

**Explanation:**

In a figure, AOC is a diameter of the circle. Using the property that, diameter subtends an angle 90° to the circle.

i.e., ∠ABC = 90°

In ΔACB

∵ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + 50° = 180°

⇒ ∠A = 180° – 140°

= 40°

∵ OA is perpendicular on AT

∴ ∠OAT = 90°

⇒ ∠OAB + ∠BAT = 90°

⇒ ∠BAT = 90° – 40°

= 50°

**4. From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PA and PB to the circle are drawn. Then, the area of the quadrilatral PAOB is**

**(a) 60 cm ^{2}**

**(b) 65 cm ^{2}**

**(c) 30 cm ^{2}**

**(d) 32.5 cm ^{2}**

**Sol.** (a) 60 cm^{2}

**Explanation:**

Here, we draw a circle of radius 5 cm having centre O. A pair of tangents AP and AB are drawn.

Join O, AP, B form a quadrilateral.

∵ OA ⊥ AP

(∵ AP is a tangent line)

In right ΔPAO,

OP^{2} = OA^{2} + AP^{2}

⇒ 13^{2} = 5^{2} + AP^{2}

⇒ AP^{2} = 169 – 25

= 144

⇒ AP = 12 cm

$$\text{Now, area of ΔOAP =}\frac{1}{2}×12×15\\\frac{1}{2}×12×15$$

= 30 cm^{2}

∴ Area of quadrilateral AOBP

= 2 ΔOAP

= 2 × 30 = 60 cm^{2}

**5. At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is**

**(a) 4 cm**

**(b) 5 cm**

**(c) 6 cm**

**(d) 8 cm**

**Sol.** (d) 8 cm

**Explanation:**

Firstly draw a circle of radius 5 cm having centre O. A tangent XY is drawn at point A.

A chord CD is drawn which is parallel to XY and at a distanct 8 cm.

Now, join OC.

Also, CE ⊥ CD

In right ΔOCE, EC^{2} = OC^{2} – OE^{2}

(By Pythagoras theorem)

= 5^{2} – 3^{2}

= 25 – 9 = 16

∴ EC = 4 cm

∴ Length of chord CD = 2 CE = 2 × 4

= 8 cm

**6. In figure, AT is a tangent to the circle with centrce O such that OT = 4 cm and ∠OTA = 30°. Then, AT is equal to**

**(a) 4 cm**

**(b) 2 cm**

$$\textbf{(c)}\space 2\sqrt{3}\space\textbf{cm}\\\textbf{(d)}\space4\sqrt{3}\space\textbf{cm}\\\textbf{Sol.}\space\textbf{(c)\space} 2\sqrt{3}\space\textbf{cm}$$

**Explanation:**

Line OA is perpendicular to tangent AT.

In ΔOAT,

$$\text{cos 30\degree}=\frac{\text{AT}}{\text{OT}}\\\Rarr\space\frac{\sqrt{3}}{2}=\frac{\text{AT}}{4}\\\Rarr\space\text{AT = 2}\sqrt{3}\space\text{cm.}$$

**7. In figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to**

**(a) 100°**

**(b) 80°**

**(c) 90°**

**(d) 75°**

**Sol.** (a) 100°

**Explanation:**

Given, ∠QPR = 50°,

We know any line draw from the centre to the point of contact is perpendicular.

∴ ∠OPR = 90°

⇒ ∠OPQ + ∠QPR = 90°

⇒ ∠OPQ = 90° – 50°

= 40°

In a given figure,

OP = OQ = radius of circle

∴ ∠OPQ = ∠OQP = 40°

∵ In ΔOPQ,

∠O + ∠P + ∠Q = 180°

⇒ ∠O = 180° – (40° + 40°)

= 180° – 80°

= 100°

**8. In figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to**

**(a) 25°**

**(b) 30°**

**(c) 40°**

**(d) 50°**

**Sol.** (a) 25°

**Explanation:**

Here, PA and PB are tangent lines.

∴ PA = PB

⇒ ∠PAB = ∠PAB = θ (Say)

In ΔPAB, ∠P + ∠A + ∠B = 180°

⇒ 50° + θ + θ = 180°

⇒ 2θ = 180° – 50°

= 130°

⇒ θ = 65°

Also, OA ⊥ PA

∴ ∠PAO = 90°

⇒ ∠PAB + ∠BAO = 90°

⇒ 65° + ∠BAO = 90°

⇒ ∠BAO = 90° – 65°

= 25°

**9. If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to**

$$\textbf{(a)}\space\frac{3}{2}\sqrt{3}\space\text{cm}$$

**(b) 6 cm**

**(c) 3 cm**

$$\textbf{(d)}\space 3\sqrt{3}\space\textbf{cm}\\\textbf{Sol.}\space\text{(d)}\space3\sqrt{3}\space\text{cm}$$

**Explanation:**

Here, P is an external point, a pair of tangents is drawn from a point P at an angle 60°.

Join AB,

Also, OB is a bisector line at an angle ∠APC.

∴ ∠APB = ∠CPB = 90°

Also, AB ⊥ AP

In right ΔBAP,

$$\text{tan 30° =}\frac{\text{AB}}{\text{AP}}=\frac{3}{\text{AP}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{3}{\text{AP}}\\\Rarr\space\text{AP = }3\sqrt{3}\space\text{cm}\\\text{Hence, length of each tangent is}\\3\sqrt{3}\space\text{cm}$$

**10. In figure, if PQR is the tangent to a circle at Q whose centre is O, AB is chord parallel to PR and ∠BQR = 70, then ∠AQB is equal to**

**(a) 20°**

**(b) 40°**

**(c) 35°**

**(d) 45°**

**Sol.** (b) 40°

**Explanation:**

Given, AB || PR

∴ ∠ABQ = ∠BQR = 70°

(Alternative angles)

Also, QD is perpendicular AB and QD bisects AB

∴ ΔADQ ~ ΔBDQ (Similar triangles)

∴ ∠A = ∠B = 70°

In ΔABQ,

∠A + ∠B + ∠Q = 180°

⇒ ∠Q = 180° – (70° + 70°)

= 40°

**Exercise 10.2**

• State whether the following statements are true or false. Justify your answer.

**1. If a chord AB subtends an angle of 60° at the centre of a circle, then angle between the tangents at A and B is also 60°.**

**Sol. **False. Since, a chord AB subtends an angle of 60° at the centrce of a circle.

i.e., ∠AOB = 60°

As OA = OB = radius of the circle

∴ ∠OAB = ∠OBA = 60°

The tangent at points, A and B are drawn, which intersect at C.

We know, OA ⊥ AC and OB ⊥ BC.

∴ ∠OAC = 90°, ∠OBC = 90°

⇒ ∠OAB + ∠BAC = 90°

and ∠OBA + ∠ABC = 90°

⇒ ∠BAC = 90° – 60° = 30°

and ∠CBA = 90° – 60 = 30°

In ΔABC,

∠BAC + ∠CBA + ∠ACB = 180°

⇒ ∠ACB = 180° – (30° + 30°) = 120°

**2. The length of tangent from an external point P on a circle is always greater than the radius of the circle.**

**Sol.** False. Because the length of tangent from an external point on a circle may or may not be greater than the radius of the circle.

**3. The length of tangent from an external point P on a circle with centre O is always less than OP.**

**Sol.** True. PT is a tangent drawn from external point P. Join OT.

∵ OT ⊥ PT

∴ OPT is a right triangle formed.

In right angle triangle, hyptoenuse is always greater than any two sides of the triangle.

∴ OP > PT

or PT < OP

**4. The angle between two tangents to a circle may be 0°.**

**Sol.** True. This may be possible only when both tangent lines are coincide or parallel to each other.

**5. If angle between two tangents drawn from a point P to a circle of radius a and centre O is 90°,**

$$\textbf{then OP = a}\sqrt{\textbf{2}}$$

**Sol.** True. From point P, two tangents are drawn.

Given, OT = a

Also, line OP bisects the ∠RPT,

∴ ∠TPO = ∠RPO = 45°

Also, OT ⊥ PT

In right ΔOTP,

$$\text{sin 45° =}\frac{\text{OT}}{\text{OP}}\\\Rarr\space\frac{1}{2}=\frac{a}{\text{OP}}\\\Rarr\space\text{OP}=a\sqrt{2}$$

**6. If angle between two tangents drawn from a point P to a circle of radius a and centre O is 60°,$$\textbf{then OP = a}\sqrt{3}.$$**

**Sol.** False. From point P, two tangents are drawn.

Given, OT = a

Also, line OP bisects the ∠RPT

∴ ∠TPO = ∠RPO = 30°

Also, OT ⊥ PT

In right ΔOTP,

$$\text{sin 30}\degree=\frac{\text{OT}}{\text{OP}}\\=\frac{1}{2}=\frac{a}{\text{OP}}$$

⇒ OP = 2a

**Exercise 10.3**

**1. The tangent to the circumcircle of an isosceles ΔABC at A, in which AB = AC, is parallel to BC.**

**Sol.** Let EAF be tangent to the circumcircle of ΔABC.

**To prove** EAF || BC

∠EAB = ∠ABC

Here, given AB = AC

⇒ ∠ABC = ∠ACB ...(i)

Since, ∠EAB and ∠BCA are angles in the alternate segments of chord AB.

∴ ∠EAB = ∠BCA ...(ii)

From Eqs. (i) and (ii), we get

∠EAB = ∠ABC

⇒ EAF || BC

**2. AB is a diameter of a circle and AC is its chord such that ∠BAC = 30°. If the tangent at C intersects AB extended at D, then BC = BD.**

**Sol.** Prove that, BC = BD

Join BC and OC.

Given, ∠BAC = 30°

⇒ ∠BCD = 30°

(Angles in alternate segments)

∴ ∠ACD = ∠ACO + ∠OCD

= 30° + 90°

= 120°

(∵ OC ⊥ CD and OA = OC = radius

⇒ ∠OAC = ∠OCA = 30°)

In ΔACD,

∠CAD + ∠ACD + ∠ADC = 180°

⇒ 30° + 120° + ∠ADC = 180°

⇒ ∠ADC = 180° – (30° + 120°) = 30°

Now, in ΔBCD,

∠BCD = ∠BDC = 30° ⇒ BC = BD

**3. Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.**

**Sol.** Let C_{2} and C_{1} be the two circles having same centre O. AC is a chord which touches the C_{1} at point D.

∴ Join OD.

Also, OD ⊥ AC

∴ AD = DC = 4 cm

(∵ Perpendicular line OD bisect the chord)

In right ΔAOD,

OA^{2} = AD^{2} + DO^{2}

⇒ DO^{2} = 5^{2} – 4^{2}

= 25 – 16 = 9

⇒ DO = 3 cm

Hence, radius of the inner circle is OD = 3 cm.

**4. If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO i.e., BO = 2 BC.**

**Sol.** Two tangents BD and BC are drawn from an external point B.

**To prove :** BO = 2BC

Given, ∠DBC = 120°

Join OC, OD and BO.

Since, BC and BD are tangents.

∴ OC ⊥ BC and OD ⊥ BD

In right ΔOBC,

$$\text{cos 60}\degree=\frac{\text{BC}}{\text{OB}}\\\Rarr\space\frac{1}{2}=\frac{\text{BC}}{\text{OB}}$$

⇒ OB = 2 BC

**Hence proved.**

**5. In figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD.**

**Sol.** Tangents OB and OD are drawn from an external point O to the circle C_{2}.

∴ OA = OC

Join OB and OD.

In right ΔOAB and ΔOCD,

OA = OC (Radius)

∠A = ∠C = 90°

and OB = OD

(Length of pair of tangent)

∴ ΔOAB ~ ΔOCD (Similar triangle)

∴ AB = CD **Hence proved.**

**6. In figure, common tangents AB and CD to two circles intersect at E. Prove that AB = CD.**

∴ OA = OC

Join OB and OD.

In right ΔOAB and ΔOCD,

OA = OC (Radius)

∠A = ∠C = 90°

and OB = OD

(Length of pair of tangent)

∴ ΔOAB ~ ΔOCD (Similar triangle)

∴ AB = CD **Hence proved.**

**7. A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.**

**Sol.** MN tangent is drawn from a point R on the circle. Chord PQ is drawn parallel to the tangent line MN.

A perpendicular line OR is drawn on MN.

Extended RO line intersects PQ on M which is also perpendicular to PQ.

In ΔPRM and ΔRQM,

∠PMR = ∠QMR = 90°

RM = RM (Common line)

PM = MQ (∵ OM bisect OQ)

∴ ΔPRM ~ ΔMRQ (Similar triangle)

⇒ PR = QR

We know, if two chords are equal, then its arcs are equal.

Hence, R bisects the arc PRQ.

**8. Prove that the tangents drawn at the ends of a chord of a circle make angles with the chord.**

**Sol.** To prove : ∠1 = ∠2

Let PQ is a chord of the circle. Tangents are drawn at the points R and Q.

Let P be another point on the circle.

Join PQ and PR.

Since, at point Q, there is a tangent,

∴ ∠2 = ∠P

(Angles in alternate segments are equal)

Since, at point R, there is a tangent,

∴ ∠1 = ∠P

(Angles in alternate segments are equal)

∴ ∠1 = ∠2 = ∠P **Hence proved.**

**9. Prove that a diamter AB of a circle bisects all those chords which are parallel to the tangent at the point A.**

**Sol.** Given, AB is a diameter of the circle.

A tangent is drawn form point A. Draw a chord CD parallel to the tangent MAN.

∵ CD is a chord of the circle and OA is a radius of the circle.

We know, any line perpendicular from centre to the chord it bisects the chord.

∴ Line AB bisect the all parallel chord.

**Exercise 10.4**

**1. Let s denotes the semi-perimeter of a ΔABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively. Prove that BD = s – b.**

**Sol.** A circle is inscribed in the ΔABC, which touches the BC, CA and AB.

Given, BC = a, CA = b and AB = c

By using the property, tangents are drawn from an external point to the circle are equal in length.

∴ BD = BF = x (Say)

DC = CE = y (Say)

and AE = AF = z (Say)

Now, BC + CA + AB = a + b + c

⇒ (BD + DC) + (CE + EA) + (AF + FB) = a + b + c

⇒ (x + y) + (y + z) + (z + x) = a + b + c

⇒ 2(x + y + z) = 2s

(∵ 2s = a + b + c = perimeter of ΔABC)

⇒ s = x + y + z

⇒ x = s – (y + z)

⇒ BD = s – a

(∵ a = BD + DC = y + z)

**Hence proved.**

**2. If AB is a chord of a circle which with centre, O, AOC is a diameter at AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB.**

**Sol.** Since, AC is a diameter line, so angle in semi-circle makes an angle 90°.

∴ ∠ABC = 90° (By property)

In ΔABC,

∠CAB + ∠ABC + ∠ACB = 180°

⇒ ∠CAB + ∠ACB = 180° – 90°

= 90° ...(i)

Since, diameter of a circle is perpendicular to the tangent.

∴ ∠CAT = 90°

⇒ ∠CAB + ∠BAT = 90° ...(ii)

From equations (i) and (ii), we get

∠CAB + ∠ACB = ∠CAB + ∠BAT

⇒ ∠ACB = ∠BAT **Hence proved.**

**3. Two circles with centres O and O′ of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O′P are two prependicular tangents tothe two circles.**

**Find the length of the common chord PQ.**

**Sol.** Here, two circles of radii OP = 3 cm and PO′ = 4 cm

These two circles intersect at P and Q.

Here, OP and PO′ are two tangents drawn at point P.

But these two tangents make an angle 90°.

Join OO′ and PN.

In right ΔOPO′,

(OO′)^{2} = (OP)^{2} + (PO)^{2}

(By Pythagoras theorem)

= (3)^{2} + (4)^{2} = 25

⇒ OO′ = 5 cm

Also, PN ⊥ OO′

Let ON = x, then NO′ = 5 – x

In right ΔOPN,

(OP)^{2} = (ON)^{2} + (NP)^{2}

⇒ (NP)^{2} = 3^{2} – x^{2}

= 9 – x^{2} ...(i)

and in right ΔPNO′,

(PO′)^{2} = (PN)^{2} + (NO′)^{2}

⇒ (4)^{2} = (PN)^{2} + (5 – x)^{2}

⇒ 7 + x^{2} – (25 + x^{2} – 10x) = 0

⇒ 10x = 18

⇒ x = 18

Again in right ΔOPN,

OP^{2} = (ON)^{2} + (NP)^{2}

⇒ 3^{2} = (1.8)^{2} + (NP)^{2}

⇒ (NP)^{2} = 9 – 3.24 = 5.76

⇒ (NP) = 2.4

∴ Length of common chord,

PQ = 2 PN = 2 × 2.4 = 4.8

**4. In a right ΔABC is which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC.**

**Sol.** Let O be the centre of the given circle. Suppose, the tangent at P meets BC at Q. Join BP.

**To prove:** BQ = QC

Since, ∠BPQ = ∠BAC

(Angles in alternate segments)

∵ AB is a diameter of the circle and we know angle in a semi-circle is right angle.

∴ ∠APB = 90°

⇒ ∠BPC = 90° ...(i)

Since, DABC is a right triangle, right angled at B.

∴ ∠BAC + ∠BCA = 90°

⇒ ∠BAC + ∠BCA = ∠BPC [From Eq. (i)]

⇒ ∠BAC + ∠BCA = ∠BPQ + ∠QPC

(∵ BPC = ∠BPQ + ∠OPQ)

⇒ ∠BAC + ∠BCA = ∠BAC + ∠QPC

(∵ BPQ = ∠BAC)

⇒ ∠BCA = ∠QPC

⇒ ∠QCA = ∠QPC

(∵ BCA = ∠OCA)

Now, in ΔPQC

∠QCA = ∠QPC ...(ii)

(∵ Sides opposite to equal angles are equal)

Since, tangents from an exterior point to a circle are equal in length,

∴ PQ = QB ...(iii)

From equations (ii) and (iii), we get

BQ = QC

Hence, PQ bisects BC.

**5. In figure, tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tange**

**Sol.** PQ and PR are two tangents drawn from an external point P.

∴ PQ = PR

(∵ The lengths of tangents drawn from an external point to a circle are equal)

⇒ ∠PQR = ∠QRP

$$=\frac{180\degree-30\degree}{2}$$

= 75°

Since, SR || QR

∴ ∠SRQ = ∠RQP = 75°

Also, ∠PQR = ∠QSR = 75°

(By alternate segment theorem)

In ΔQRS, ∠Q + ∠R + ∠S = 180°

⇒ ∠Q = 180° – (75° + 75°)

= 30°

∴ ∠RQS = 30°

**6. Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.**

**Sol.** Let mid-point of an arc AMB is M and let TMT is the tangent to the circle.

Join AB, AM and MB.

Since, arc AM = arc MB

⇒ Chord AM = Chord MB

In ΔAMB, AM = MB

⇒ ∠MAB = ∠MBA

(Equal sides corrresponding to the equal angles) ... (i)

Since, TMT′ is a tangent line.

∴ ∠AMT = ∠MBA

(Angles in alternate segments are equal)

= ∠MAB [From Eq. (i)]

⇒ But ∠AMT and ∠MAB are alternate angles, which is possible only when

AB || TMT′

**7. In figure, O is the centre of a circle of radius 5 cm, T is a point such that OT = 13 cm andOT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB.**

**Sol.** Given, OT = 13 cm and OP = 5 cm

Since, OP ⊥ PT

In right ΔOPT,

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

(By Pythagoras theorem)

⇒ PT^{2} = (13)^{2} – (5)^{2}

= 169 – 25 = 144

⇒ PT = 12 cm

∵ OT = 13

⇒ OE + ET = 13

⇒ ET = 13 – 5 = 8

[∵ OE = 5 cm (radius)]

Let AP = x

⇒ AT = 12 – x

Since, AB is tangent at E.

∴ OE ⊥ AE

In right ΔAET,

(AT)^{2} = (AE)^{2} + (ET)^{2}

(∵ The length of tangents are drawn from an external point to a circle are equal)

PA = AE = x

⇒ (AT)^{2} = (PA)^{2} + (ET)^{2}

⇒ (12 – x)^{2} = (x)^{2} + (8)^{2}

144 + x^{2} – 24x = x^{2} + 64

⇒ 24x = 80

$$\Rarr\space x=\frac{10}{3}\\\therefore\space\text{AB = 2AE = 2x}\\=2×\frac{10}{3}=\frac{20}{3}\text{cm}$$

**8. The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA = 110°, find ∠CBA.**

**Sol.** Here, AB is a diameter of the circle from point C and a tangent is drawn which meets at a point P.

Join OC,

Here, OC is radius.

∴ OC ⊥ PC

Now, ∠PCA = 110° (Given)

⇒ ∠PCO + ∠OCA = 110°

⇒ 90° + ∠OCA = 110°

⇒ ∠OCA = 20°

∴ OC = OA = radius of circle

⇒ ∠OCA = ∠OAC = 20°

(∵ Two sides are equal, then their opposite angles are equal)

Since, PC is a tangent.

∴ ∠BCP = ∠CAB = 20°

(Angles in a alternate segments are equal)

In ΔPBC,

∠BPC + ∠PCB + ∠PBC =180°

∵ In ΔPCA,

∠P + ∠C + ∠A = 180°

∠P = 180° – ( ∠C + ∠A)

= 180° – (110° + 20°)

= 180° – 130° = 50°

⇒ 50° + 20° + ∠PBC = 180°

⇒ ∠PBC = 110°

∵ APB is a straight loine.

∴ ∠PBC + ∠CBA = 180°

⇒ ∠CBA = 180° – 110°

= 70°

**9. If an isosceles ΔABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.**

**Sol.** In a circle, ΔABC is inscribed.

Join OB, OC and OA.

Since, BC is a chord of the circle.

∴ OA is perpendicular bisector of BC.

Let AM = x, then OM = 9 – x

(∵ OA = radius = 9 cm)

In right ΔAMC,

AC^{2} = AM^{2} + MC^{2}

(By Pythagoras theorem)

⇒ MC^{2} = 6^{2} – x^{2} ...(i)

and in right ΔOMC,

OC^{2} = OM^{2} + MC^{2}

(By Pythagoras theorem)

⇒ MC^{2} = 9^{2} – (9)

(By Pythagoras theorem)

From equations (i) and (ii), we get

6^{2} – x^{2} = 9^{2} – (9 – x)^{2}

⇒ 36 – x^{2} = 81 – (81 + x^{2} – 18x)

⇒ 36 = + 18x ⇒ x = 2

∴ AM = x = 2

In right ΔABM,

AB^{2} = BM^{2} + AM^{2}

(By Pythagoras theorem)

6^{2} = BM^{2} + 2^{2}

⇒ BM^{2} = 36 – 4 = 32

$$\Rarr\space\text{BM}=4\sqrt{2}\\\therefore\text{BC = 2 BM = 8}\sqrt{2}\text{cm}\\\therefore\text{Area of ΔABC =}\frac{1}{2}×\text{Base}×\text{Height}\\=\frac{1}{2}×\text{BC}×\text{AM}\\=\frac{1}{2}×8\sqrt{2}×2\\=8\sqrt{2}\text{cm}^{2}$$