**Question:**

AP and PQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. If OA = 15 cm, then AP + AQ =

(a) 12 cm

(b) 18 cm

(c) 24 cm

(d) 36 cm

**Solution:**

Let us first put the given data in the form of a diagram.

We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore,

$O P \perp A P$

$A P^{2}=O A^{2}-O P^{2}$

$A P^{2}=15^{2}-9^{2}$

$A P^{2}=225-81$

$A P^{2}=144$

$A P=\sqrt{144}$

$A P=12$

Since tangents drawn from an external point will be equal in length,

*AP = AQ*

Since, *AP* = 12

*AQ* = 12

*AP + AQ* = 12 + 12

*AP + AQ* = 24

Therefore option (c) is the correct answer to this question.

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