Find the equation of the polar of the point (2, -1) with respect to the circle $x^{2} + y^{2} - 3 x + 4 y - 8 = 0$

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Solution

Given circle is $x^{2} + y^{2} - 3 x + 4 y - 8 = 0$ ...................(i)
Given point is (2, -1) let P = (2, -1) Now equation of the polar of point P with respect to circle (i)
$x \times 2 + y \times (- 1) - 3 (\frac{x + 2}{2}) + 4 (\frac{y - 1}{2}) - 8 = 0$
or 4x - 2y - 3x - 6 + 4y - 4 - 16 = 0 or x +2y - 26 = 0

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What is the equation of polar of the point (1, 2) with respect to the circle $x^{2} + y^{2} + 8 x + 2 y + 1 = 0$

The equation of the circle concentric with $x^{2} - 3 x + 4 y - c = 0$ and passing through (-1, -2) is

x^{2} + y^{2} + 3 x + 4 y + 1 = 0

(- 1, - 2)

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