The circle S=0 cuts the circles C_1=x^2+y^2-8 | vedclass

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(B) Let the equation of circle $S$ be $x^2+y^2+2gx+2fy+c=0$. Since $S$ cuts $C_1$ and $C_2$ orthogonally,we have: $2g(-4) + 2f(-1) = c + 16 \implies -

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$\left(\frac{11}{3}, \frac{-7}{6}\right)$

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(B) Let the equation of circle $S$ be $x^2+y^2+2gx+2fy+c=0$. Since $S$ cuts $C_1$ and $C_2$ orthogonally,we have: $2g(-4) + 2f(-1) = c + 16 \implies -8g - 2f = c + 16 \quad (i)$ $2g(-2) + 2f(-2) = c - 1 \implies -4g - 4f = c - 1 \quad (ii)$ Subtracting $(ii)$ from $(i)$: $-4g + 2f = 17 \implies 2f = 4g + 17 \implies f = 2g + 8.5$. The common chord of $S=0$ and $C_1=0$ is $S - C_1 = 0$: $(2g+8)x + (2f+2)y + (c-16) = 0$. Comparing this with the given chord $2x + 13y - 15 = 0$: $\frac{2g+8}{2} = \frac{2f+2}{13} = \frac{c-16}{-15} = k$. $2g+8 = 2k \implies g = k-4$. $2f+2 = 13k \implies f = \frac{13k-2}{2}$. Substitute into $2f = 4g + 17$: $13k-2 = 4(k-4) + 17 \implies 13k-2 = 4k-16+17 \implies 9k = 3 \implies k = \frac{1}{3}$. Then $g = \frac{1}{3} - 4 = \frac{-11}{3}$ and $f = \frac{13(1/3)-2}{2} = \frac{7/3}{2} = \frac{7}{6}$. The centre of $S$ is $(-g, -f) = \left(\frac{11}{3}, \frac{-7}{6}\right)$.

List-$I$	List-$II$
$(A)$ Point circles of $S_\alpha=0$	$(i)$ do not exist
$(B)$ Point circles of $S_\beta=0$	(ii) intersecting
$(C)$ The circles in $S_\alpha=0$ are	(iii) non-intersecting
$(D)$ The circles in $S_\beta=0$ are	(iv) $(\pm \sqrt{k}, 0)$
	$(v)$ $(0, \pm \sqrt{k})$

The circle $S=0$ cuts the circles $C_1=x^2+y^2-8x-2y+16=0$ and $C_2=x^2+y^2-4x-4y-1=0$ orthogonally. If the common chord of $S=0$ and $C_1=0$ is $2x+13y-15=0$,then the centre of $S=0$ is

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