A circle C1 passes through the origin O and has diameter 4 on the positive x-axis. The line y = 2x gives a chord OA of a circle C1. Let C2 be the circle with OA as a diameter. If the tangent to C2 at the point A meets the x-axis at P and y-axis at Q, then QA : AP is equal to :

C1 : x2+y2−4x=0tan θ=2C2 is a circle with OA as diameter.So, tangent at A on C2 is perpendicular to ORLet OA=l∴QAAP=ℓ cot θℓ tan θ=1tan2 θ=14

A circle C1 passes through the origin O and has diameter 4 on the positive x-axis. The line y = 2x gives a chord OA of a circle C1. Let C2 be the circle with OA as a diameter. If the tangent to C2 at the point A meets the x-axis at P and y-axis at Q, then QA : AP is equal to :

C1 : x2+y2−4x=0tan θ=2C2 is a circle with OA as diameter.So, tangent at A on C2 is perpendicular to ORLet OA=l∴QAAP=ℓ cot θℓ tan θ=1tan2 θ=14

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A circle C₁ passes through the origin O and has diameter 4 on the positive x-axis. The line y = 2x gives a chord OA of a circle C₁. Let C₂ be the circle with OA as a diameter. If the tangent to C₂ at the point A meets the x-axis at P and y-axis at Q, then QA : AP is equal to :

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$\begin{array}{l} C_{1} : x^{2} + y^{2} - 4 x = 0 \\ \tan θ = 2 \end{array}$

C₂ is a circle with OA as diameter.

So, tangent at A on C₂ is perpendicular to OR

Let $OA = l$

$\begin{array}{l} ∴ \frac{QA}{AP} = \frac{ℓ \cot θ}{ℓ \tan θ} \\ = \frac{1}{\tan^{2} θ} = \frac{1}{4} \end{array}$

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A circle C1 passes through the origin O and has diameter 4 on the positive x-axis. The line y = 2x gives a chord OA of a circle C1. Let C2 be the circle with OA as a diameter. If the tangent to C2 at the point A meets the x-axis at P and y-axis at Q, then QA : AP is equal to :

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A circle C₁ passes through the origin O and has diameter 4 on the positive x-axis. The line y = 2x gives a chord OA of a circle C₁. Let C₂ be the circle with OA as a diameter. If the tangent to C₂ at the point A meets the x-axis at P and y-axis at Q, then QA : AP is equal to :