SEBA Class 10 Maths Chapter 10 Circles

SEBA Class 10 Maths Chapter 10 Circles Solutions English Medium As Per New Syllabus. to each chapter is provided in the list so that you can easily browse through different chapters SEBA Class 10 Maths Chapter 10 Circles Notes and select need one. SEBA Class 10 Maths Chapter 10 Circles Question Answers Download PDF. SEBA Class 10 General Mathematics Solutions.

SEBA Class 10 Maths Chapter 10 Circles

Also, you can read the SEBA book online in these sections Solutions by Expert Teachers as per Board of Secondary Education, Assam (SEBA) Book guidelines. SEBA Class 10 Maths Chapter 10 Circles are part of All Subject Solutions. Here we have given SEBA Solutions For Class 10 Maths for All Chapters, You can practice these here.

Circles

Chapter: 10

Exercise 10.1

1. How many tangents can a circle have?

Ans: A circle can have infinitely many tangents.

2. Fill in the blanks:

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

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

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

Ans: A line intersecting a circle in two points is called a secant.

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

Ans: 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 _________.

Ans: The common point of a tangent to a circle and the circle is called point of contact.

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

(a) 12cm

(b) 13cm

(c) 8.5 cm

(d) √119

Ans: (d) √119

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

Ans: 

A circle having centre O is drawn. Here, I is the given line.

AB is a secant such that AB || l. 

PT is a tangent such that PT || l.

5. In figure 10.6, the length of the tangent AP, from an external point A is 24 cm. If the distance of the point A from the centre O of the circle is 25cm, then the diameter of the circle is

(A) 15 cm

(B) 14 cm

(C) 7 cm

(D) 12 cm

Ans: (B) 14 cm

6. In figure 10.7, two circle with their respective centres intersect each other at A and B. AB intersects OO’ at M, then ∠OMA is

(A) 50°

(B) 60°

(C) 180°

(D) 90°

Sol”. (D) 90°

7. Assertion (A): The length of the diameter of the circle with centre O is 6 cm. PT is a tangent from an external point P to the circle at the pt. T. Distance of point P and the centre O is 5cm. Then the length of the tangent PT is 4 cm. (fig. 10.8)

Reason (R): A tangent to a circle is perpendicular to the radius through the point of contact.

(A) Assertion (A) and Reason (R) both are true, Reason (R) is the correct explanation for Assertion (A).

(B) Assertion (A) and Reason (R) both are true, Reason (R) is not the correct explanation for Assertion

(C) Assertion (A) is true, but Reason (R) is false.

(D) Assertion (A) is false, but Reason (R) is true.

Ans: (A) Assertion (A) and Reason (R) both are true, Reason (R) is the correct explanation for Assertion (A).

8. The line segment joining the points of contact of two parallel tangents to a circle is a/an.

(i) radius of the circle.

(ii) diameter of the circle.

(iii) secant of the circle.

(iv) arc of the circle.

Which of the above are correct?

(A) Only (ii)

(B) (i) and (ii)

(C) Only (iii)

(D) (i), (ii) and (iv)

Ans: (A) Only (ii)

9. If a chord of a circle is equal to its radius then the angle subtended by the chord at the centre of the circle is

(a) 30°

(b) 60°

(c) 45°

(d) 90°

Ans: (b) 60°

10. If PT is tangent drawn from P to a circle touching it at T and O is the centre of the circle. Find the sum of the measures of ∠OPT and ∠POT

Ans: 

OT ⊥ PT

(Radius is perpendicular to the tangent at the point

contact)

∴  ∠OTP = 90°

∠OTP + ∠OPT + ∠POT = 180° (for ΔPOT)

⇒ 90° + ∠OPT + ∠POT = 180°

⇒ ∠OPT + ∠POT = 90°

∴ Sum of the measure of ∠OPT and ∠POT = 90°

Exercise 10.2

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 25cm. The radius of the circle is

(a) 7cm 

(b) 12cm 

(c) 15cm 

(d) 24.5cm

Ans: (a) 7cm 

2. In fig. 10.15, 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

Ans: (b) 70º

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 equal to

(a) 50 

(b) 60º

(c) 70º

(d) 80º

Ans: 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 = 360

∠OAP + ∠APB + ∠PBO + ∠BOA = 360

90º + 80º + 90º + ∠BOA = 360

∠BOA = 100

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.

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

Ans:

Let AB be the 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 = 90º ∠OAS 

= 90º

∠OBP = 90º

∠OBQ = 90º

It can be observed that 

∠OAR = ∠OBQ (Alternate interior angles)

∠OAS = ∠(Alternate interior angles) 

Since alternate interior angles)

Since alternative interior angles are equal, lines PQ and RS Will parallel.

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

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.

Ans: 

Let us consider a circle centred at point O.

AB is a tangent drawn on this circle from point A.

Given that,

OA = 5cm and AB = 4cm 

In ∆ABO,

OB ⊥ AB ( radius ⊥ tangent at at the point of Contact) 

Applying pythagoras theorem in ∆ABO,

OB 丄 AB (radius  丄 tangent at the points of contract )

Applying pythagoras theorem in ∆ABO, we obtain 

AB² + BO² = OA²

4² + BO² = 25  

16 + BO² = 25

BO² = 3 

Hence, 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.

Ans: 

Let the two concentric circles be a contrast at point O. and let PQ be the chord of the larger circle which touches the smaller circle at point. 

Therefore PQ is tangent to the smaller circle at the point. 

OA 丄 PQ (As oa is the circle)

Applying pythagoras theorem in ∆OAP, we obtain)

OA ⊥ PQ ( As AP² = OP² )

3² + AP² = 5²

AP² = 16 

AP = 4

In ∆OPQ,

Since OA ⊥ PQ 

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

∴ PQ = 2AP = 2 × 4 = 8 

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