The pole of $3 x + 4 y - 45 = 0$ with respect to the circle $x^{2} + y^{2} - 6 x - 8 y + 5 = 0$ is

(6, 8)

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(8, 6)

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(4, 3)

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(3, 4)

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Solution

The correct option is A $(6, 8)$
Let the pole be $(h, k)$ , so the equation of polar is
$T = 0 \Rightarrow h x + k y - 3 (x + h) - 4 (y + k) + 5 = 0 \Rightarrow (h - 3) x + (k - 4) y - (3 h + 4 k - 5) = 0$
Given equation of polar is
$3 x + 4 y - 45 = 0$
Comparing both the equations, we get
$\frac{h - 3}{3} = \frac{k - 4}{4} = \frac{3 h + 4 k - 5}{45}$
Now,
$\frac{h - 3}{3} = \frac{k - 4}{4} \Rightarrow h = 3 + \frac{3 k - 12}{4} = \frac{3 k}{4} \dots (1)$
Also,
$\frac{k - 4}{4} = \frac{3 h + 4 k - 5}{45} \Rightarrow 45 k - 180 = 12 h + 16 k - 20$
Using equation $(1)$ , we get
$\Rightarrow 20 k = 160 \Rightarrow k = 8 \Rightarrow h = 6$

Hence, the pole is $(6, 8)$

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