NCERT Solutions Class 10 Maths Chapter 10 Circles Exercise 10.2

NCERT Solutions Class 10 Maths Chapter 10 Exercise 10.2 will help you understand the concepts of circles, especially concepts related to tangents. This exercise extends the foundational concepts of the circles and tangents studied in earlier exercises. The concepts covered here will also form the basis for more complex topics for future studies of mathematics. The solutions provided here are aligned with the latest CBSE syllabus and examination preparation. So, let’s begin by diving into this essential topic that is vital for Class 10 Maths.

1.0Download NCERT Solutions Class 10 Maths Chapter 10 Circles Exercise 10.2 : Free PDF

2.0Introduction to Tangents of Circles

A tangent to a circle is a line that is straight and touches the circle at only one point. The definition is crucial when learning about circles and is important in many geometric problems. This exercise will help extend our knowledge of tangents and the number of times they can be drawn from points in different positions with respect to the circle, whether the point is inside the circle, lies on the circle, or is outside the circle. 

3.0Class 10 Maths Chapter 10 Circles Exercise 10.2: Key Concepts

Number of Tangents from Different Positions

The exercise begins by examining how many tangents can be drawn from a point based on its position relative to the circle.

Case 1: Point Inside the Circle

In the first case, when the point is provided inside the circle, you can see that it is not possible to construct a tangent through it. This is because any line drawn from an interior point of a circle will cut the circle at two points. This does not satisfy the key characteristic of a tangent, it touches a circle at only one point.

Case 2: Point on the Circle

When a point lies on the circumference of the circle, only one tangent can be drawn through this point. This point is the point of tangency on the circle's edge and is perpendicular to the circle's radius. The point is also known as the point of contact of a tangent.

Case 3: Point Outside the Circle 

It is the most interesting case of the tangents to a circle. In this, a point lies on the outside of the circle, and the number of tangents that can be drawn from this point is exactly two. The tangents formed from this point touch the circle at two different points. In this exercise, this property of external points gives rise to an interesting theorem, which is used in the exercise many times for solving complex problems.  

The Length of Tangents from an External Point:

If a point is outside the circle, then the two tangents from the exterior point to the circle are always of equal length. This gives us a basic theorem that establishes the equality of these tangent lengths. For example, let point P be given to the external point of a circle with centre O. From point P, two tangents meet the circle at points A and B. According to this theorem, AP = BP. The theorem is proved with the help of concepts of congruency of the triangles. 

By working through NCERT Solutions Class 10 Maths Chapter 10 Exercise 10.2, you will be able to develop a better understanding of tangents and circles, ultimately helping to develop a solid foundation for more complicated concepts. So, keep practising, and watch your ability improve!

4.0NCERT Class 10 Maths Chapter 10 Circles Exercise 10.2 : Detailed Solutions

1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. The radius of the circle is -

(A) 7 cm

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

Solution:

From figure,

r² = (25)² - (24)²

= 625 - 576 = 49

r = 7 cm

Hence, the correct option is (A).

2. In the 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°

Solution:

TQ and TP are tangents to a circle with centre O and ∠POQ = 110°.

Therefore, OP ⊥ PT and OQ ⊥ QT.

This implies ∠OPT = 90° and ∠OQT = 90°.

Now, in the quadrilateral TPOQ, we get:

∠PTQ + 90° + 110° + 90° = 360° (Angle sum property of a quadrilateral)

∠PTQ + 290° = 360°

∠PTQ = 360° - 290° = 70°

Hence, the correct option is (B).

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

(A) 50°

(B) 60°

(C) 70°

(D) 80°

Solution:

In the figure,

ΔOAP ≅ ΔOBP (SSS congruence)

⇒ ∠POA = ∠POB = (1/2)∠AOB  ...(i)

Also, ∠AOB + ∠APB = 180°

⇒ ∠AOB + 80° = 180°

⇒ ∠AOB = 100° ...(ii)

Then, from (i) and (ii),

∠POA = (1/2) × 100° = 50°

Hence, the correct option is (A).

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

Solution:

In the figure, PQ is the diameter of the given circle and O is its center.

Let tangents AB and CD be drawn at the end points of the diameter PQ.

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

Therefore, PQ ⊥ AB

=> ∠APQ = 90° and PQ ⊥ CD

=> ∠PQD = 90°

=> ∠APQ = ∠PQD

But they form a pair of alternate angles.

Therefore, AB || CD.

Hence, the two tangents are parallel.

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

Solution:

Let us assume a circle with centre O and let AB be the tangent intersecting the circle at point P.

Also let us assume a point X such that XP is perpendicular to AB.

We have to prove that XP passes through centre O.

We know that

The tangent of a circle is perpendicular to the radius at the point of contact.

=> OP ⊥ AB (Theorem-1)

So, ∠OPB = 90°

We have already assumed that XP is perpendicular to AB.

∠XPB = 90°

Now from equation (1) and (2)

∠OPB = ∠XPB = 90°

This condition is possible only if line XP passes through O. Since, XP passes through centre O.

Therefore, it is proved that the perpendicular at the point of contact of the tangent of a circle passes through the centre.

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

Solution:

The tangent to a circle is perpendicular to the radius through the point of contact.

Therefore, ∠OTA = 90°.

Now, in the right triangle OTA, we have:

OA² = OT² + AT² [Pythagoras theorem]

=> 5² = OT² + 4²

=> OT² = 5² - 4²

=> OT² = (5 - 4)(5 + 4)

=> OT² = 1 × 9 = 9 = 3²

=> OT = 3

Thus, the 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.

Solution:

In the figure, the two concentric circles have their centre at O. The radius of the larger circle is 5 cm and that of the smaller circle is 3 cm.

AB is a chord of the larger circle and it touches the smaller circle at P.

Join OA, OB, and OP.

Now, OA = OB = 5 cm,

OP = 3 cm.

And OP ⊥ AB,

i.e., ∠OPA = ∠OPB = 90°.

⇒ ΔOAP ≅ ΔOBP (RHS congruence)

⇒ AP = BP = (1/2)AB or AB = 2AP

By Pythagoras theorem,

OA² = AP² + OP²

⇒ (5)² = AP² + (3)²

⇒ AP² = 25 - 9 = 16

⇒ AP = 4 cm

⇒ AB = 2 × 4 cm = 8 cm

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

Solution:

In the given figure, we observe that:

AP = AS  (Since AP and AS are tangents to the circle drawn from point A)  ...(i)

Similarly, 

BP = BQ  ...(ii)

CR = CQ  ...(iii)

DR = DS  ...(iv)

Adding equations (i), (ii), (iii), and (iv), we get:

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

This simplifies to:

AB + CD = AD + BC

Note: This is known as Pitot's theorem.

9. In the given figure, XY and X'Y' are two parallel tangents to a circle with center O, and another tangent AB with point of contact C intersects XY at A and X'Y' at B. Prove that ∠AOB = 90°.

Solution:

In the given figure, join OC. We have triangles AOP and AOC for which:

AP = AC (Both are tangents from point A)

OP = OC (Each is a radius)

OA = OA (Common side)

Therefore, triangle AOP is congruent to triangle AOC (by SSS congruence).

This implies that angle PAO = angle CAO.

Hence, angle PAC = 2 x angle OAC.  (Equation i)

Similarly, angle QBC = 2 x angle OBC. (Equation ii)

Adding equations (i) and (ii), we get:

angle PAC + angle QBC = 2 x (angle OAC + angle OBC)

Since the sum of co-interior angles is 180 degrees, we have:

180 degrees = 2 x (angle OAC + angle OBC)

Therefore, angle OAC + angle OBC = (1/2) * 180 degrees = 90 degrees. (Equation iii)

Now, in triangle AOB, we have:

angle AOB + angle OAC + angle OBC = 180 degrees

Substituting the value from equation (iii), we get:

angle AOB + 90 degrees = 180 degrees

Therefore, angle AOB = 90 degrees.

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.

Solution: 

Let PA and PB be two tangents drawn from an external point P to a circle with centre O.

Now, in right triangle OAP and right triangle OBP, we have:

PA = PB [Tangents to a circle from an external point]

OA = OB [Radii of the same circle]

OP = OP [Common]

Triangle OAP is congruent to triangle OBP [By SSS congruency]

Therefore, angle OPA = angle OPB [By C.P.C.T.]

and angle AOP = angle BOP

=> angle APB = 2 * angle OPA and angle AOB = 2 * angle AOP

But angle AOP = 90° - angle OPA

=> 2 * angle AOP = 180° - 2 * angle OPA

=> angle AOB = 180° - angle APB

=> angle AOB + angle APB = 180°

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

Solution

Let ABCD be a parallelogram such that its sides touch a circle with center O.

=> AP = AS [Tangents from an external point are equal]

=> BP = BQ [Tangents from an external point are equal]

=> CR = CQ [Tangents from an external point are equal]

=> DR = DS [Tangents from an external point are equal]

Adding these equations,

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

=> AB + CD = AD + BC

=> 2AB = 2BC

=> AB = BC

and AB = DC

=> AB = BC = CD = DA

=> ABCD is a rhombus.

Hence proved.

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:

In the given figure, BD = 8 cm and DC = 6 cm.

Then, we have BE = 8 cm (since BE = BD)

and CF = 6 cm (since CF = CD)

Suppose AE = AF = x cm.

In triangle ABC,

a = BC = 6 cm + 8 cm = 14 cm

b = CA = (x + 6) cm

c = AB = (x + 8) cm

s = (a + b + c) / 2 = [14 + (x + 6) + (x + 8)] / 2 cm

= (2x + 28) / 2 cm

= (x + 14) cm

Area of triangle ABC = √[s(s - a)(s - b)(s - c)]

= √[(x + 14) × x × 8 × 6]

= √[48x × (x + 14)] cm²  ...(i)

Also, area of triangle ABC = area of triangle OBC + area of triangle OCA + area of triangle OAB

= (1/2) × 4 × a + (1/2) × 4 × b + (1/2) × 4 × c

= 2(a + b + c) = 2 × 2s = 4s

= 4(x + 14) cm²  ...(ii)

From (i) and (ii),

√[48x × (x + 14)] = 4 × (x + 14)

On squaring both sides,

48x × (x + 14) = 16 × (x + 14)²

3x = x + 14

x = 7 cm

Then, AB = c = (x + 8) cm = (7 + 8) cm = 15 cm

and AC = b = (x + 6) cm = (7 + 6) cm = 13 cm

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

Solution:

Let ABCD be a quadrilateral circumscribing a circle with center O.

Join AO, BO, CO, and DO.

From the figure, ∠DAO = ∠BAO.

[Since, AB and AD are tangents]

Let ∠DAO = ∠BAO = ∠1.

Also, ∠ABO = ∠CBO.

[Since, BA and BC are tangents]

Let ∠ABO = ∠CBO = ∠2.

Similarly, we can follow the same reasoning for vertices C and D, assigning angles ∠3 and ∠4 accordingly.

Recall that the sum of the angles in quadrilateral ABCD is 360°.

Therefore:

∠DAO + ∠BAO + ∠ABO + ∠CBO + ... (angles at C and D) = 360°

This can be rewritten as:

2(∠1 + ∠2 + ∠3 + ∠4) = 360°

Dividing by 2 gives:

∠1 + ∠2 + ∠3 + ∠4 = 180°

In triangle AOB, ∠BOA = 180° - (∠1 + ∠2).

In triangle COD, ∠COD = 180° - (∠3 + ∠4).

Adding these two equations:

∠BOA + ∠COD = 360° - (∠1 + ∠2 + ∠3 + ∠4)

Substituting the value of (∠1 + ∠2 + ∠3 + ∠4) = 180°:

∠BOA + ∠COD = 360° - 180° = 180°

Therefore, AB and CD subtend supplementary angles at O.

Since the sum of the angles at the center is 360°, it follows that AD and BC also subtend supplementary angles at O.

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

5.0Benefits of Studying NCERT Solutions Class 10 Maths Chapter 10 Circles Exercise 10.2

• Helps in understanding the properties of tangents and their applications.
• Builds a strong foundation for entrance exams like NTSE and Olympiads.
• Easy-to-follow solutions that would help students learn independently.
• Well-structured answers make revision quick and efficient.
•  Covers important questions frequently asked in CBSE board exams.