If $3 x + 5 y + 17 = 0$ is polar for the circle $x^{2} + y^{2} + 4 x + 6 y + 9 = 0$ , then the pole is

(- 1, 2)

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(1, 2)

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(1, - 2)

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(2, 1)

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Solution

The correct option is B $(1, 2)$
Given circle is
$x^{2} + y^{2} + 4 x + 6 y + 9 = 0$
Let pole be $(h, k)$
Now, the equation of polar,
$T = 0 \Rightarrow x (h) + y (k) + 2 (x + h) + 3 (y + k) + 9 = 0$
$\Rightarrow (h + 2) x + (k + 3) y + (2 h + 3 k + 9) = 0$
But, the given polar equation is
$3 x + 5 y + 17 = 0$
Comparing both, we get
$\frac{h + 2}{3} = \frac{k + 3}{5} = \frac{2 h + 3 k + 9}{17}$
Now,
$\frac{h + 2}{3} = \frac{k + 3}{5} \Rightarrow 5 h + 10 = 3 k + 9 \Rightarrow h = \frac{3 k - 1}{5} \dots (1)$
And
$\frac{k + 3}{5} = \frac{2 h + 3 k + 9}{17} \Rightarrow 17 k + 51 = 10 h + 15 k + 45$
Using equation $(1)$ , we get
$\Rightarrow 2 k + 6 = 6 k - 2 \Rightarrow k = 2 \Rightarrow h = 1$

Hence, the pole is $(1, 2)$

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