If the inverse of $P (- 3, 5)$ w.r.t to a circle is $(1, 3)$ then polar of $P$ w.r.t to the circle is

x + 2 y = 7

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2 x - 2 y + 11 = 0

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2 x - y + 1 = 0

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2 x - y - 1 = 0

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Solution

The correct option is C $x + 2 y = 7$
Given points are P(-3,5) and Q(1,3)

then Its equation is
$\frac{y - 5}{x + 3} = \frac{3 - 5}{1 + 3} \to \frac{y - 5}{x + 3} = - \frac{1}{2}$

$\to 2 y - 10 = - x - 3 \to x + 2 y - 7 = 0$

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The point of contact of the circle $x^{2} + y^{2} + 2 x + 2 y + 1 = 0$ and $x^{2} + y^{2} - 2 x + 2 y + 1 = 0$ is

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