The equation of the pair of transverse common | vedclass

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(C) Given circles are $S_1 \equiv x^2 + y^2 + 2x + 2y + 1 = 0$ and $S_2 \equiv x^2 + y^2 - 2x - 2y + 1 = 0$. Centre of $S_1$ is $C_1 = (-1, -1)$ and r

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$xy = 0$

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(C) Given circles are $S_1 \equiv x^2 + y^2 + 2x + 2y + 1 = 0$ and $S_2 \equiv x^2 + y^2 - 2x - 2y + 1 = 0$. Centre of $S_1$ is $C_1 = (-1, -1)$ and radius $r_1 = \sqrt{(-1)^2 + (-1)^2 - 1} = 1$. Centre of $S_2$ is $C_2 = (1, 1)$ and radius $r_2 = \sqrt{(1)^2 + (1)^2 - 1} = 1$. The distance between centres $C_1 C_2 = \sqrt{(1 - (-1))^2 + (1 - (-1))^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$. Since $C_1 C_2 > r_1 + r_2$ $(2\sqrt{2} > 2)$,the circles are separate. The transverse common tangents intersect at the internal centre of similitude $P$,which divides $C_1 C_2$ in the ratio $r_1 : r_2 = 1 : 1$. $P = \left( \frac{1(-1) + 1(1)}{1+1}, \frac{1(-1) + 1(1)}{1+1} \right) = (0, 0)$. The pair of tangents from $(0, 0)$ to $S_1$ is given by $T^2 = S_1 S_{11}$,where $T = x(0) + y(0) + 1(x+0) + 1(y+0) + 1 = x + y + 1$ and $S_{11} = 0^2 + 0^2 + 2(0) + 2(0) + 1 = 1$. Thus,$(x + y + 1)^2 = (x^2 + y^2 + 2x + 2y + 1)(1)$. $x^2 + y^2 + 1 + 2xy + 2x + 2y = x^2 + y^2 + 2x + 2y + 1$. $2xy = 0 \Rightarrow xy = 0$.

The equation of the pair of transverse common tangents drawn to the circles $x^2 + y^2 + 2x + 2y + 1 = 0$ and $x^2 + y^2 - 2x - 2y + 1 = 0$ is

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