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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x^{2} + y^{2} - 2 x - 4 y + 3 = 0

x^{2} + y^{2} - 2 x + 4 y + λ = 0

x^{2} + y^{2} - 2 x + 4 y + 1 = 0

x^{2} + y^{2} - 2 x + 4 y + 3 = 0

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x^{2} + y^{2} + 2 y - 3 = 0,

x^{2} + y^{2} - 3 x + 4 y - 8 = 0

(x - 2 y + 3) + λ (2 x - 3 y + 4) = 0 (λ ϵ R)

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