Circle

Solution of NCERT Exercise 10.2 (Part1)

Question (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 25 cm. The radius of the circle is

(A) 7 cm

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

Answer: (A) 7 cm

Explanation

Here Q is the given point.

According to question,

PQ = 24 cm

OQ = 5 cm

∴ Radius = OP = ?

Now, because, OP is radius and PQ is tangent of the circle

Thus, ∠ OPQ = 90⁰

Thus, according to Pythagoras theorem

OQ² = OP² + PQ²

⇒ 25² = OP² + 24²

⇒ 625 = OP² + 576

⇒ OP² = 625 – 576

⇒ OP² = 49

`:. OP = sqrt(49)`

⇒ OP = 7 cm

Thus, option (A) 7 cm is the correct answer.

Question (2) In Figure , if TP and TQ are the two tangents to a circle with center O so that ∠110⁰, then ∠PTQ is equal to

(A) 60⁰

(B) 70⁰

(C) 80⁰

(D) 90⁰

Answer : (B) 70⁰

Explanation

Given, TP and TQ are tangents to the given circle with center O

And, OP and OQ are radius of the given circle

Therefore, TP ⊥ OP

And, TP ⊥ OQ

∴ ∠ OPT = ∠ OQT = 90⁰

Now, OPTQ forms a quadrilateral.

We know that sum of all interior angles of a quadrilateral = 360⁰

∴ ∠ OPT + ∠ OQT + ∠ POQ + ∠ PTQ = 360⁰

⇒ 90⁰ + 90⁰ + 110⁰ + ∠ PTQ = 360⁰

⇒ 290⁰ + ∠ PTQ = 360⁰

⇒ ∠ PTQ = 360⁰ – 290⁰

⇒ ∠ PTQ = 70⁰

Thus, option (B) 70⁰ is the correct answer.

Question (3) If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle 80⁰, then ∠ POA is equal to

(A) 50⁰

(B) 60⁰

(C) 70⁰

(D) 80⁰

Answer: (A) 50⁰

Explanation:

Let, PA and PB are tangents to the given circle

And OA and OB are radius of the given circle

∴ PA ⊥ OA

And PB ⊥ OB

∴ ∠OAP = 90⁰

And ∠ OBP = 90⁰

Now, AOBP form a quadrilateral.

We know that sum of all internal angles of a quadrilateral is equal to 360⁰

∴ ∠OAP + ∠ OBP + ∠ APB + ∠ AOB = 360⁰

⇒ 90⁰ + 90⁰ + 80⁰ + ∠ AOB = 360⁰

⇒ 260⁰ + ∠ AOB = 360⁰

⇒ ∠ AOB = 360⁰ – 260⁰

⇒ ∠ ABO = 100⁰

Now, Join O and P

In triangle, OAP and OBP

OA = OB

[∵ OA and OB are radii of same circle]

And PA = PB

[∵ PA and PB are tangent to same circle from same point P]

And OP = OP

[Common side in both of the triangles]

∴ According to SSS congruence criterion

Δ OAP = Δ OBP

∴ ∠ AOP = ? ∠ AOB

⇒ ∠ AOP = 1/2 × 100 ⁰

⇒ ∠ AOP = 50⁰

Thus, option (A) 50⁰ is the correct answer.

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

Solution:

Let there is a circle with center O.

Let AB is the diameter of the given circle.

Let RS and PQ are two tangents at the two ends of diameter of the given circle.

Thus, it is to prove that RS is parallel to PQ

Since, RS is a tangent at opint A and OA is the radius of the circle,

∴ OA ⊥ RS

∴ ∠ OAR = 90⁰

And, ∠ OAS = 90⁰

Similarly, OB is another radius of the same circle and PQ is the tangent to the point B

Thus, OB ⊥ PQ

And ∠ OBP = OBQ = 90⁰

Now, ∠ OAR = ∠ OBQ = 90⁰[Alternate interior angles]

And ∠ OAS = ∠ OBP = 90⁰ [Alternate interior angles]

Since, alternate interior angles of RS and PQ are equal,

Thus, RS is parallel to PQ.

Thus, tangents drawn at the ends of a diameter of a circle are parallel. Proved

Question (5) Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.

Solution:

Let AB is a tangent to the circle with center O.

And P is the point of contact of tangent AB to the circle.

Thus, it is to prove that perpendicular OP to the tangent passes through the center O of the circle.

This can be proved using contradiction by assuming that perpendicular at the point of contact to the tangent to a circle does not pass through the center.

Let perpendicular to the tangent AB at point of touch P does not passes through the center O of the circle.

Rather, let the perpendicular at point P to AB passes through an other point O' which is other than the center O.

Since, OP' is perpendicular to AP

&ther4; ∠ O'PB = 90⁰

And we know that the line joining the center and point of contact of a tangent, i.e. radius of the circle is perpendicular to the tangent.

&there; OP ⊥ AB

∴ ∠ OPB = 90⁰

This menas line OP and O'P should be conincide.

But OP and O'P will coincide only if O and O' will coincide.

But, it assumed that O' is other point than O, which is not possible, i.e. our assumption is incorrect.

Thus, it is proved that the perpendicular at the point of contact to the tangent to a circle passes through the center.

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

Solution:

Given, OA = 5 cm

AB = 4 cm

Thus, Radius OB =?

Since, ∠ B = 90⁰

Thus, According to Pythagoras Theorem

OA² = OB² + AB²

⇒ 5² = OB² + 4²

⇒ 25 = OB² + 16

⇒ OB² = 25 – 16

⇒ OB² = 9

∴ OB = √ 9

⇒ OB = 3 cm Answer

MCQs Test

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