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A circle $C_{1}$ passes through the origin O and has diameter 4 on the positive x-axis. The line $y = 2 x$ gives a chord OA of a circle $C_{1}$ . Let $C_{2}$ be the circle with OA as a diameter. If the tangent to $C_{2}$ at the point A meets the x-axis at P and y-axis at Q, then $Q A : A P =$

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1:3

1:5

1:4

2:5

answer is C.

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Detailed Solution

Let $θ$ Inclination of $\bar{OA}$

Equation of OA is $y = 2 x \Rightarrow \tan θ = 2$

$C_{2}$ is a circle with OA as diameter.

So the tangent at A on $C_{2} ⊥ \bar{OA}$

Let $OA = ℓ, \underline{OQA} = θ$

Now $QA = OAcot θ = ℓ \cot θ$

$AP = OAtan θ = ℓ \tan θ$

$∴ \frac{QA}{AP} = \frac{1}{{Tan}^{2} θ} = \frac{1}{4} = 1 : 4$

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