The point of contact of the circles x^2+y^2+2 | vedclass

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(B) The given equations of the circles are:$S_1 \equiv x^2+y^2+2x+2y+1=0$$S_2 \equiv x^2+y^2-2x+2y+1=0$For circle $S_1$,the centre is $C_1 = (-1, -1)$

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$(0, -1)$

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(B) The given equations of the circles are:$S_1 \equiv x^2+y^2+2x+2y+1=0$$S_2 \equiv x^2+y^2-2x+2y+1=0$For circle $S_1$,the centre is $C_1 = (-1, -1)$ and the radius is $r_1 = \sqrt{(-1)^2 + (-1)^2 - 1} = \sqrt{1+1-1} = 1$.For circle $S_2$,the centre is $C_2 = (1, -1)$ and the radius is $r_2 = \sqrt{(1)^2 + (-1)^2 - 1} = \sqrt{1+1-1} = 1$.The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(1 - (-1))^2 + (-1 - (-1))^2} = \sqrt{2^2 + 0^2} = 2$.Since $d = r_1 + r_2$ $(2 = 1 + 1)$,the circles touch each other externally.The point of contact is the midpoint of the line segment joining the centres $C_1$ and $C_2$:$	ext{Point of contact} = \left(\frac{-1+1}{2}, \frac{-1-1}{2}ight) = (0, -1)$.

The point of contact of the circles $x^2+y^2+2x+2y+1=0$ and $x^2+y^2-2x+2y+1=0$ is

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