Circles MCQ: Introduction & Explanation

Ans: d) Infinite

Explanation: A tangent to a circle is defined as a straight line which touches the circle at one point, called the point of tangency. At the point of tangency, the tangent of the circle is perpendicular to the radius drawn to the point of contact.

Hence a circle can have any number of tangents on its surface so that it can be drawn perpendicular to the centre of the circle as shown in the below diagram.

Ans: c) Always greater than OA

Explanation: As we know that the tangent is perpendicular to the radius of the circle, then the angle comes out to be 90°. Then the opposite side to that angle is the hypotenuse which is OA, for the right-angled triangle OAB. For any right-angled triangle, the hypotenuse is the only longest side of the triangle. Therefore, the length of the tangent drawn from an external point is always less than the line joining the point and the centre of the circle.

Ans: (c) 50°

Explanation:

According to the given question, we can draw a diagram for the circle having tangent from the external point T.

We know that angle in Semicircle is right angle hence,

∠ABC = 90°

In the triangle Δ ACB,

⇒ ∠A + ∠B + ∠C = 180

⇒ ∠A = 180° – (90° + 50°)

⇒ ∠A = 40°

Or ∠OAB = 40°

Therefore, ∠BAT = 90° – 40° = 50°

Ans: b) 70°

Explanation: According to the given information in the question, we can draw that the OP is the radius of the circle to the tangent PT and OQ is the radius to the tangents drawn to the tangents TQ.

So, OP ⊥ PT and TQ ⊥ OQ (perpendicular lines)

∴ ∠OPT = ∠OQT = 90°

Now, in the quadrilateral 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° = 180°+ 110°+ ∠PTQ = 360°

⇒ ∠PTQ = 70°

Ans: (a) 3 cm

Explanation: According to the given information in the question, we can draw that the OB is the perpendicular line from the centre and joining A,B gives the tangent to the circle.

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

So, OB ⊥ AB

Given, OA = 5cm and AB = 4 cm

Now, In the Δ ABO,

OA² = AB² + BO² (by Using Pythagoras theorem)

⇒ 52 = 42 + BO²

⇒ BO² = 25 – 16

⇒ BO² = 9

⇒ BO = 3.

As the value of the length of the triangle can not be negative, we consider only the positive value of BO.

Ans: (c) 5 cm

Explanation: Join the points on the circle to the centre of the circle. Join OP and OQ.

Tangents AP = AQ

In triangle Δ APO and Δ AQO,

AP = AQ since the radius of the circle is uniform.

OP = OQ (radius of same circle)

Thus, ΔAPO ~ ΔAQO. (Similar triangles)

The Quadrilateral POQA is a square

OP = OQ = AP = AQ

So, AP = AQ = 5 Cm

And AP = OP (Proved)

Therefore, radius AP = OP = OQ = 5 cm

Ans: (a) 8 cm

Explanation: According to the given information in the question, we can draw concentric circles with common centre O.

From the above figure, AB can be identified as the tangent to the smaller circle at the point P.

∴ OP ⊥ AB

By applying Pythagoras theorem, in triangle Δ OPA

⇒ OA ²= AP ² + OP ²

⇒ 5 ² = AP ² + 32

⇒ AP ² = 25 – 9

⇒ AP = 4

Now, as OP ⊥ AB,

Since the perpendicular from the circle’s centre bisects the chord, AP will be equal to PB

So, AB = 2 × AP = 2 × 4 = 8 cm.

Ans: (b) R₁² + R₂² = R²

Explanation: consider two circles with radii R₁ and R₂,

The area of a circle with radius r is given by πr².

Given that the sum of areas is equal to others then

⇒ A₁+A₂ = A

⇒π (R₁² + R₂²) = πR²

⇒R₁² + R₂² = R²

Hence the sum of squares of the radius is equal to the other circle area.

Ans: (b) Area of the circle > Area of the square

Explanation: According to the given parameters

The circumference of the circle is 2πr

The circumference of the square of side a is given by 4a.

∴ From the given condition, we have 2π r = 4a

(22/7) r = 2a

⇒ 11r = 7a

⇒ a = (11/7)a

⇒ r = (7/11)a

The Area of circle = A1 = πr2²

The Area of square = A2 = a²

we have

A1 = π × (7/11)² (From above equation)

= (22/7) × (49/121) a²

= (14/11) a2 and A2 = a²

∴ A1 = (14/11) A2

⇒ A1 > A2

Therefore, Area of the circle > Area of the square.

Ans: (a) 19%

Explanation: Let the radius of the circle be “r”.

So, the area of a circle = πr²

If the radius of a circle is diminished by 10%, new radius, r = r – (10% of r)

⇒r = 90% of radius

⇒r = (90/100) r

⇒r = 9r/10

We know that the area of a circle is πr2 square units.

Now, substitute r = 9r/10

⇒A = π(9r/10) ²

⇒A = π(81r2/100)

Therefore, the change in area = πr2 – π(81r2/100)

⇒A = (19/100)πr²

⇒A = 0.19 πr²

Hence, the area is diminished by 19%.