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(D) The family of circles passing through the intersection of $S_1: x^2+y^2-2x-3=0$ and $S_2: x^2+y^2-2y=0$ is given by $S_1 + \lambda S_2 = 0$ for $\

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$2x^2+2y^2-2x-2y-3=0$

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(D) The family of circles passing through the intersection of $S_1: x^2+y^2-2x-3=0$ and $S_2: x^2+y^2-2y=0$ is given by $S_1 + \lambda S_2 = 0$ for $\lambda 
eq -1$. $(1+\lambda)x^2 + (1+\lambda)y^2 - 2x - 2\lambda y - 3 = 0$. Dividing by $(1+\lambda)$,we get $x^2 + y^2 - \frac{2}{1+\lambda}x - \frac{2\lambda}{1+\lambda}y - \frac{3}{1+\lambda} = 0$. The center is $C = \left(\frac{1}{1+\lambda}, \frac{\lambda}{1+\lambda}ight)$ and radius $r = \sqrt{\frac{1}{(1+\lambda)^2} + \frac{\lambda^2}{(1+\lambda)^2} + \frac{3}{1+\lambda}} = \sqrt{\frac{1+\lambda^2+3+3\lambda}{(1+\lambda)^2}}$. Since $x+y+1=0$ is a tangent,the perpendicular distance from $C$ to the line equals $r$: $\frac{|\frac{1}{1+\lambda} + \frac{\lambda}{1+\lambda} + 1|}{\sqrt{1^2+1^2}} = r$ $\Rightarrow \frac{|\frac{1+\lambda+1+\lambda}{1+\lambda}|}{\sqrt{2}} = r$ $\Rightarrow \frac{|\frac{2(1+\lambda)}{1+\lambda}|}{\sqrt{2}} = r$ $\Rightarrow \sqrt{2} = r$. Squaring both sides: $2 = \frac{4+3\lambda+\lambda^2}{(1+\lambda)^2}$ $\Rightarrow 2(1+2\lambda+\lambda^2) = 4+3\lambda+\lambda^2$ $\Rightarrow 2+4\lambda+2\lambda^2 = 4+3\lambda+\lambda^2$ $\Rightarrow \lambda^2+\lambda-2=0$. $(\lambda+2)(\lambda-1)=0$,so $\lambda=1$ or $\lambda=-2$. For $\lambda=1$: $2x^2+2y^2-2x-2y-3=0$. For $\lambda=-2$: $-x^2-y^2-2x+4y-3=0 \Rightarrow x^2+y^2+2x-4y+3=0$ (not in options). Thus,the correct equation is $2x^2+2y^2-2x-2y-3=0$.

$A$ circle $S$ passes through the points of intersection of the circles $x^2+y^2-2x-3=0$ and $x^2+y^2-2y=0$. If $x+y+1=0$ is a tangent to the circle $S$,then the equation of $S$ is

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