$P$ is in the exterior of a circle at distance $34$ from the centre $O$ . A line through $P$ touches the circle at $Q$ . $P Q = 16$ , find the diameter of the circle.

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Solution

REF.image
AP is tangent to circle
$\Rightarrow A P ⊥ O A (O A = r a d i u s)$
In $Δ$ QPO, by Pythagoras theorem,
$O P^{2} = O Q^{2} + P Q^{2}$
$\Rightarrow O Q^{2} = O P^{2} - P Q^{2}$
$\Rightarrow O Q^{2} = 50 \times 18 = 25 \times 36$
$\Rightarrow O Q = 5 \times 6 = 30 c m$
$\Rightarrow r a d i u s = 30 c m \Rightarrow d i a m e t e r = 60 c m .$

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P is in the exterior of a circle at distance 34 from the centre O. A line through P touches the circle at Q. PQ=16, find the diameter of the circle.

REF.imageAP is tangent to circle ⇒AP⊥OA(OA=radius)In Δ QPO, by Pythagoras theorem,OP2=OQ2+PQ2⇒OQ2=OP2−PQ2⇒OQ2=50×18=25×36⇒OQ=5×6=30cm⇒radius=30cm⇒diameter=60cm.

$P$ is in the exterior of a circle at distance $34$ from the centre $O$ . A line through $P$ touches the circle at $Q$ . $P Q = 16$ , find the diameter of the circle.