The locus of the centers of the circles that | vedclass

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(A) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2-1=0$ and $S_2: x^2+y^2-2x+y=0$ is given by $S_1 + kS_2 = 0

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a line whose equation is $x+2y=0$.

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(A) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2-1=0$ and $S_2: x^2+y^2-2x+y=0$ is given by $S_1 + kS_2 = 0$ for $k 
eq -1$. $(x^2+y^2-1) + k(x^2+y^2-2x+y) = 0$ $(1+k)x^2 + (1+k)y^2 - 2kx + ky - (1+k) = 0$ Dividing by $(1+k)$,we get $x^2 + y^2 - \frac{2k}{1+k}x + \frac{k}{1+k}y - 1 = 0$. The center $(h, k')$ of this circle is given by $(-g, -f) = \left(\frac{k}{1+k}, \frac{-k}{2(1+k)}ight)$. Let $x = \frac{k}{1+k}$ and $y = \frac{-k}{2(1+k)}$. Then $2y = \frac{-k}{1+k}$. Adding $x$ and $2y$,we get $x + 2y = \frac{k}{1+k} - \frac{k}{1+k} = 0$. Thus,the locus is the line $x+2y=0$.

The locus of the centers of the circles that are passing through the intersection of the circles $x^2+y^2=1$ and $x^2+y^2-2x+y=0$ is

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