Find the equation of the circle which passes | vedclass

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(C) Let $S_1 = x^2+y^2-8x-6y+21=0$ and $S_2 = x^2+y^2-2x-15=0$. Using the family of circles equation $S_1 + \lambda S_2 = 0$,we have: $(x^2+y^2-8x-6y+

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$3(x^2+y^2)-18x-12y+27=0$

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(C) Let $S_1 = x^2+y^2-8x-6y+21=0$ and $S_2 = x^2+y^2-2x-15=0$. Using the family of circles equation $S_1 + \lambda S_2 = 0$,we have: $(x^2+y^2-8x-6y+21) + \lambda(x^2+y^2-2x-15) = 0$. Since the circle passes through the point $(1, 2)$,substitute $x=1$ and $y=2$ into the equation: $(1^2+2^2-8(1)-6(2)+21) + \lambda(1^2+2^2-2(1)-15) = 0$. $(1+4-8-12+21) + \lambda(1+4-2-15) = 0$. $6 + \lambda(-12) = 0$. $12\lambda = 6 \Rightarrow \lambda = \frac{1}{2}$. Substituting $\lambda = \frac{1}{2}$ back into the family equation: $(x^2+y^2-8x-6y+21) + \frac{1}{2}(x^2+y^2-2x-15) = 0$. Multiply the entire equation by $2$: $2(x^2+y^2-8x-6y+21) + (x^2+y^2-2x-15) = 0$. $2x^2+2y^2-16x-12y+42 + x^2+y^2-2x-15 = 0$. $3x^2+3y^2-18x-12y+27 = 0$. $3(x^2+y^2)-18x-12y+27 = 0$.

List-$I$	List-$II$
$A$. $x^2+y^2+2x+8y-23=0$,$x^2+y^2-4x-10y+19=0$	$I$. $0$
$B$. $x^2+y^2=1$,$x^2+y^2-2x-6y+6=0$	$II$. $1$
$C$. $x^2+y^2-8x+2y=0$,$x^2+y^2-2x-16y+25=0$	$III$. $2$
$D$. $x^2+y^2=4$,$x^2+y^2-2x=0$	$IV$. $3$
	$V$. $4$

Find the equation of the circle which passes through the point $(1, 2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0$.

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