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Q.

State true or false.

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. Here, ∠ AOB = 90°.

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a

True

b

False

answer is A.

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Detailed Solution

Given that 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.
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, is intersecting XY at A and

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, is intersecting XY at A and

Join OC.
OC is perpendicular to AB (radius of the circle is perpendicular to its tangent)
Therefore,

∠ A C O = ∠ B C O = 90 °

In

Δ A O P a n d Δ A O C

AO = AO (Common side)

O P = O C

(radius)

A P = A C

(the length of tangents drawn from external point are equal)
Therefore,

Δ A O P ≅ Δ A O C

⇒ ∠ A O P = ∠ A O C

… (1)
In

Δ B O C a n d Δ B O Q

OB = OB (common side)

O C = O Q

(radius)

B C = B Q

(Theorem 2)
Therefore,

Δ B O C ≅ Δ B O Q

⇒ ∠ B O C = ∠ B O Q

.… (2)
For line PQ,

∠ A O P + ∠ A O C + ∠ B O C + ∠ B O Q = 180 °

Substitute the equations (1) and (2) in the above equation,
we get,

⇒ ∠ A O C + ∠ A O C + ∠ B O C + ∠ B O C = 180 ° ⇒ 2 ∠ A O C + 2 ∠ B O C = 180 ° ⇒ 2 (∠ A O C + ∠ B O C) = 180 °

⇒ ∠ A O C + ∠ B O C = 180 ° 2 ⇒ ∠ A O B = 90 °

Hence,

∠ A O B = 90 °

Therefore, the given statement is true.
Hence, option (1) is correct.

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