Find the pole of the straight line $a x + b y + 3 a^{2} + 3 b^{2} = 0$ with respect to the circle $x^{2} + y^{2} + 2 a x + 2 b y = a^{2} + b^{2} .$

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Solution

Polar of the point $(h, k)$ with respect to the circle $x^{2} + y^{2} + 2 a x + 2 b y = a^{2} + b^{2}$ is given by $x h + y k + a x + a h + b y + b k = a^{2} + b^{2}$
Comparing the above equation with the equation of the straight line $a x + b y + 3 a^{2} + 3 b^{2} = 0$ , we have
$\frac{h + a}{a} = \frac{k + b}{b} = \frac{a h + b k - a^{2} - b^{2}}{3 a^{2} + 3 b^{2}}$
$\Rightarrow b (h + a) = a (k + b)$
$\Rightarrow b h = a k . . . (1)$
Also, $3 (a^{2} + b^{2}) (h + a) = a (a h + b k - a^{2} - b^{2})$ or $3 a^{2} h + 3 a^{3} + 3 b^{2} h + 3 a b^{2} = a^{2} h + a b k - a^{3} - a b^{2}$
$\Rightarrow 2 a^{2} h + 4 a^{3} + 3 b^{2} h + 4 a b^{2} - a b k = 0$ ...... $(2)$
Substituting equation $(1)$ in equation $(2)$ , we have
$(2 a^{2} + 3 b^{2}) h + 4 a^{3} + 4 a b^{2} = b^{2} h$
$∴ h (2 a^{2} + 2 b^{2}) = - 4 a (a^{2} + b^{2})$
$∴ h = - 2 a$
Also, $k = \frac{b h}{a} = \frac{b \times (- 2 a)}{a} = - 2 b$
The pole is thus $(- 2 a, - 2 b)$

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