What is the polar of the point (2, 3) with respect to the circle $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$

x + y = 10

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x + y = 12

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x -y = 6

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x - y = 10

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Solution

The correct option is B
x + y = 12

$[x + y = 12]$

This can be found by the formula T = 0

Given equation is $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$

i.e., $2 x + 3 y - (x + 2) - 2 (y + 3) - 4 = 0$

$x + y - 2 - 6 - 4 = 0$

$x + y = 12$

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