NCERT Solutions Class 10 Maths Chapter 10 Circles

Question

Our NCERT Solutions Class 10 Maths Chapter 10 Circles is very helpful in studying for the CBSE board examination. These solutions are based on the latest syllabus proposed by the CBSE. Prepared and designed by mentors who have expertise in the subject matter. Mentors have designed the solutions to all the queries with the help of graphs, equations, diagrams, formulas, shortcuts, tips, and tricks. These NCERT solutions are a great way in understanding difficult concepts, solve tough questions, revise the concepts, complete your homework, and do your assignments. This NCERT Solution is a perfect study material for clearing the basics of the tough chapter like circle very easily as well as preparing for competitive exams like JEE Advance, JEE Mains, NTSE, Olympiad. It is suggested by the experts to go through the solutions to clear the concepts and score good marks in the CBSE board exam.

Our NCERT Solutions Class 10 Maths Chapter 10 Circles have discussed all the concepts in detail including in-text questions and exercise questions. In this chapter, we have discussed chapter 10 circles. Important topics discussed here are:

• Circle – A circle is a locus of all the points which are equidistant from a given point. The formula for the circle is (x-h)² + (y-k)² = r²
• Drawing a circle.
•  Parts of a circle –
  • Annulus – the area bounded by who concentric circle.
  • Arc – it is a connected curve of a circle.
  • Sector – an area bounded by two radii and an arc.
  • Segment – an area bounded by a chord and an arc of a circle is known as a segment. A segment does not have a center.
  • Centre – Midpoint of a circle.
  • Chord – a line whose endpoints lie on the circumference of the circle.
  • Diameter – a line segment whose endpoints lie on the circumference of the circle and also passed through the center of the circle.
  • Radius – the shortest distance between the center and circumference of the circle.
  • Secant – a straight line that cuts the circumference of the circle at two points. it is an extended chord.
  • Tangent of a circle – tangent is a line that touches the circle at a single point and tangency is the point where the tangent meets the circle. A tangent is always perpendicular to the radius of the circle.

In the NCERT Solutions Class 10 Maths, we have designed our solutions in such a way that it aids the learning and revision of a student in the least possible time. Ample practice of the questions discussed here is the best way to gain a good command of the related topics.

Get started now to ace your examination.     

Answer 1

NCERT Solutions Class 10 Maths Chapter 10 Circles

1. How many tangents can a circle have?

Answer:

A circle can have infinite number of tangents because a circle have infinite number of points on it and at every point a tangent can be drawn.

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) The common point of a tangent to a circle and the circle is called …………

Answer:

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

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

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

(iv) The common point of a tangent to a circle and the circle is called the 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

(D) √119 cm

Answer:

In △OPQ, angle P is right angle.       [Since radius is perpendicular to tangent]

Using Pythagoras theorem, OQ² = PQ² + OP²

Hence, the option (D) is correct. 

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

Answer:

Consider a circle with centre O, Let Ab is the given line.

Now draw a perpendicular from O to line AB, which intersect AB at P.

Now take two points on the line PO, one at circle X and another Y inside the circle. Draw lines parallel to AB and passing through X and Y. CD and EF are the required lines.

5. 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:

Let O be the centre of the circle. 

Given that, 

OQ = 25cm and PQ = 24 cm 

As the radius is perpendicular to the tangent at the point of contact, 

Therefore, OP ⊥ PQ 

Applying Pythagoras theorem in ∆OPQ, we obtain 

OP² + PQ² = OQ² 

OP² + 24² = 25² 

OP² = 625 − 576 

OP² = 49 

OP = 7 

Therefore, the radius of the circle is 7 cm. 

Hence, alternative (A) is correct

6. In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110º, then ∠PTQ is equal to 

(A) 60º

(B) 70º

(C) 80º

(D) 90º

Answer:

It is given that TP and TQ are tangents. 

Therefore, radius drawn to these tangents will be perpendicular to the tangents. 

Thus, OP ⊥ TP and OQ ⊥ TQ 

∠OPT = 90º 

∠OQT = 90º 

In quadrilateral POQT,

Sum of all interior angles = 360º

∠OPT + ∠POQ +∠OQT + ∠PTQ = 360º

⇒ 90º + 110º + 90º + ∠PTQ = 360º

⇒ ∠PTQ = 70º

Hence, alternative (B) is correct

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

(A) 50º

(B) 60º

(C) 70º

(D) 80º

Answer:

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents. 

Thus, OA ⊥ PA and OB ⊥ PB 

∠OBP = 90º and ∠OAP = 90º 

In AOBP,

Sum of all interior angles = 360º 

∠OAP + ∠APB +∠PBO + ∠BOA = 360º

90º + 80º + 90º + ∠BOA = 360º

∠BOA = 100º

In ∆OPB and ∆OPA,

AP = BP (Tangents from a point) 

OA = OB (Radii of the circle) 

OP = OP (Common side) 

Therefore, ∆OPB ≅ ∆OPA (SSS congruence criterion)

And thus, ∠POB = ∠POA

Hence, alternative (A) is correct.

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

Answer:

Let AB is a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.

Radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⟂ RS and OB ⟂ PQ

∠OAR = 90° ∠OAS

= 90°

∠OBP = 90°

∠OBQ = 90°

It can be observed that

∠OAR = ∠OBQ (Alternate interior angles)

∠OAS = ∠OBP (Alternate interior angles)

Since alternate interior angles are equal, lines PQ and RS will be parallel.

Answer 2

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

Answer:

Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

We have to prove that the line perpendicular to AB at P passes through centre O. 

We shall prove this by contradiction method. 

Let us assume that the perpendicular to AB at P does not pass through centre O. 

Let it pass through another point O’. Join OP and O’P.

As perpendicular to AB at P passes through O',therefore,

∠O'PB = 90  ..................(1)

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

\(\therefore\) ∠OPB = 90  ..................(2)

Comparing equations (1) and (2), we obtain 

∠O'PB = ∠OPB .................(3)

From the figure, it can be observed that,

∠O'PB < ∠OPB .................(4)

Therefore, ∠O'PB < ∠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.

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.

Answer:

Here, AB is the tangent that is drawn on the circle from a point A.

So, the radius OB will be perpendicular to AB i.e. OB ⊥ AB

We know, OA = 5cm and AB = 4 cm

Now, In △ABO,

OA² =AB²+BO² (Using Pythagoras theorem)

5² = 4²+BO²

BO² = 25-16

BO² = 9

BO = 3

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

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.

Answer:

Let the two concentric circles be centered at point O. And let PQ be the chord of the larger circle which touches the smaller circle at point A. Therefore, PQ is tangent to the smaller circle.

OA ⟂ PQ (As OA is the radius of the circle)

Applying Pythagoras theorem in △OAP, we obtain

In △OPQ,

Since OA ⟂ PQ,

AP = AQ (Perpendicular from the center of the circle bisects the chord)

\(\therefore\) PQ  = 2AP = 2 x 4 = 8

Therefore, the length of the chord of the larger circle is 8 cm.

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

Answer:

It can be observed that 

DR = DS (Tangents on the circle from point D) ………….. (1)

CR = CQ (Tangents on the circle from point C) …………… (2)

BP = BQ (Tangents on the circle from point B) …………… (3)

AP = AS (Tangents on the circle from point A) …………… (4)

Adding all these equations, we obtain 

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

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

CD + AB = AD + BC

13. In the given 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°.

Answer:

Let us join point O to C.

Now the triangles △OPA and △OCA are similar using SSS congruency as:

(i) OP = OC They are the radii of the same circle

(ii) AO = AO It is the common side

(iii) AP = AC These are the tangents from point A

So, △OPA ≅ △OCA

Similarly,

△OQB ≅ △OCB

So,

∠POA = ∠COA ….... (i)

And, ∠QOB = ∠COB ….... (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 equation (ii) we get,

2∠COA+2∠COB = 180°

∠COA+∠COB = 90°

∴∠AOB = 90°

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.

Answer:

Let us consider a circle centered at point O. 

Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle. 

It can be observed that 

OA (radius) ⊥ PA (tangent) 

Therefore, ∠OAP = 90° 

Similarly, OB (radius) ⊥ PB (tangent) 

∠OBP = 90° 

In quadrilateral OAPB, 

Sum of all interior angles = 360º 

∠OAP +∠APB+∠PBO +∠BOA = 360º 

90º + ∠APB + 90º + ∠BOA = 360º 

∠APB + ∠BOA = 180º 

Hence, it can be observed 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.

Answer 3

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

Answer:

Since ABCD is a parallelogram, 

AB = CD ……………(1) 

BC = AD …………..(2)

It can be observed that 

DR = DS (Tangents on the circle from point D) 

CR = CQ (Tangents on the circle from point C) 

BP = BQ (Tangents on the circle from point B) 

AP = AS (Tangents on the circle from point A) 

Adding all these equations, we obtain 

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

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

CD + AB = AD + BC 

On putting the values of equations (1) and (2) in this equation, we obtain 

2AB = 2BC 

AB = BC ……………(3) 

Comparing equations (1), (2), and (3), we obtain 

AB = BC = CD = DA 

Hence, ABCD is a rhombus.

16. 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 given figure). Find the sides AB and AC.

Answer:

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be x. 

In ∆ABC, 

CF = CD = 6cm (Tangents on the circle from point C) 

BE = BD = 8cm (Tangents on the circle from point B) 

AE = AF = x (Tangents on the circle from point A) 

AB = AE + EB = x + 8 

BC = BD + DC = 8 + 6 = 14 

CA = CF + FA = 6 + x 

2s = AB + BC + CA

2s = x + 8 + 14 + 6 + x

2s = 28 + 2x 

s = 14 + x

Area of ∆ABC = Area of ∆OBC + Area of ∆OCA + Area of ∆OAB

Either x + 14 = 0 or x − 7 =0 

Therefore, x = −14 and 7

However, x = −14 is not possible as the length of the sides will be negative. 

Therefore, x = 7 

Hence, 

AB = x + 8 = 7 + 8 = 15 cm 

CA = 6 + x = 6 + 7 = 13 cm

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

Answer:

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. 

Let us join the vertices of the quadrilateral ABCD to the center of the circle.

Consider ∆OAP and ∆OAS, 

AP = AS (Tangents from the same point) 

OP = OS (Radii of the same circle) 

OA = OA (Common side) 

∆OAP ≅ ∆OAS (SSS congruence criterion)

thus, ∠POA = ∠AOS 

∠1 = ∠8 

Similarly,

∠2 = ∠3 

∠4 = ∠5 

∠6 = ∠7 

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

(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º 

2∠1 + 2∠2 + 2∠5 + 2∠6 = 360º 

2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º 

(∠1 + ∠2) + (∠5 + ∠6) = 180º 

∠AOB + ∠COD = 180º 

Similarly, we can prove that BOC + DOA = 180º

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