Class 10 Maths MCQs for Chapter 10 (Circles) are provided here online with solved answers. These objective questions will help students to score good marks in the board exam. These multiple choice questions are prepared chapter-wise as per the latest CBSE syllabus and NCERT textbook.

## Class 10 Maths MCQs for Circles

Solve the multiple-choice problems where each question carries 4 options, among which one is correct. Practice them and verify your answers with the given answers here. Also, find important questions for class 10 Maths here.

**1. A circle has a number of tangents equal to**

(a)0

(b)1

(c)2

(d)Infinite

Answer: d

Explanation: A circle has infinitely many tangents, touching the circle at infinite points on its circumference.

**2. A tangent intersects the circle at:**

(a)One point

(b)Two distinct point

(c)At the circle

(d)None of the above

Answer: a

Explanation: A tangent touches the circle only on its boundary and do not cross through it.

**3. A circle can have _____parallel tangents at a single time.**

(a)One

(b)Two

(c)Three

(d)Four

Answer: b

Explanation: A circle can have two parallel tangents at the most.

**4. If the angle between two radii of a circle is 110º, then the angle between the tangents at the ends of the radii is:**

(a)90º

(b)50º

(c)70º

(c)40º

Answer: c

Explanation: If the angle between two radii of a circle is 110º, then the angle between tangents is 180º − 110º = 70º. (By circles and tangents properties)

**5. The length of the tangent from an external point A on a circle with centre O is**

(a)always greater than OA

(b)equal to OA

(c)always less than OA

(d)Cannot be estimated

Answer: (c)

Explanation: Since the tangent is perpendicular to the radius of the circle, then the angle between them is 90º. Thus, OA is the hypotenuse for the right triangle OAB right-angled at B. As we know, for any right triangle, the hypotenuse is the longest side, therefore the length of the tangent from external point is always less than the OA.

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

(a)65°

(b)60°

(c)50°

(d)40°

Answer: c

Explanation: As per the given question:

∠ABC = 90 (Angle in Semicircle)

In ∆ACB

∠A + ∠B + ∠C = 180°

∠A = 180° – (90° + 50°)

∠A = 40°

Or ∠OAB = 40°

Therefore, ∠BAT = 90° – 40° = 50°

**7. 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: b

Explanation: As per the given question:

We can see, OP is the radius of the circle to the tangent PT and OQ is the radius to the tangents TQ.

So, OP ⊥ PT and TQ ⊥ OQ

∴ ∠OPT = ∠OQT = 90°

Now, in the quad. POQT, we know that the sum of the interior angles is 360°

So, ∠PTQ + ∠POQ + ∠OPT + ∠OQT = 360°

Now, by putting the respective values we get,

⇒ ∠PTQ + 90° + 110° + 90° = 360°

⇒ ∠PTQ = 70°

**8. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. The radius of the circle is:**

(a)3cm

(b)5cm

(c)7cm

(d)10cm

Answer: 3

Explanation: As per the given question:

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

So, OB ⊥ AB

Given, OA = 5cm and AB = 4 cm

Now, In △ABO,

OA^{2} = AB^{2} + BO^{2} (Using Pythagoras theorem)

⇒ 5^{2} = 4^{2} + BO^{2}

⇒ BO^{2} = 25 – 16

⇒ BO^{2} = 9

⇒ BO = 3

**9. If a parallelogram circumscribes a circle, then it is a:**

(a)Square

(b)Rectangle

(c)Rhombus

(d)None of the above

Answer: c

**10. Two concentric circles are of radii 5 cm and 3 cm. The length of the chord of the larger circle which touches the smaller circle is:**

(a)8

(b)10

(c)12

(d)18

Answer: a

Explanation: As per the given question:

From the above figure, AB is tangent to the smaller circle at point P.

∴ OP ⊥ AB

By Pythagoras theorem, in triangle OPA

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

⇒ 5^{2} = AP^{2} + 3^{2}

⇒ AP^{2} = 25 – 9

⇒ AP = 4

Now, as OP ⊥ AB,

Since the perpendicular from the center of the circle bisects the chord, AP will be equal to PB

So, AB = 2AP = 2 × 4 = 8 cm