The pole of a straight line with repect to the circle $x^{2} + y^{2} = a^{2}$ lies on the circle $x^{2} + y^{2} = 9 a^{2} .$ Prove that the straight line touches the circle $x^{2} + y^{2} = a^{2} / 9.$

x^{2} + y^{2} = a^{2}

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x^{2} + y^{2} = 4 a^{2}

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x^{2} + y^{2} = \frac{a^{2}}{9}

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x^{2} + y^{2} = \frac{a^{2}}{81}

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Solution

The correct option is C $x^{2} + y^{2} = \frac{a^{2}}{9}$
let the pole be $(3 a cos θ, 3 a sin θ)$
Then equation of polar w.r.t $x^{2} + y^{2} = a^{2}$ is $3 a x cos θ + 3 a y sin θ = a^{2}$
$\Rightarrow 3 x cos θ + 3 y sin θ = a$
perpendicular distance from origin to $3 x cos θ + 3 y sin θ = a$ is $\frac{a^{2}}{9}$

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