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(B) The family of circles passing through the intersection of two circles ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} + \lambda {S_2} = 0$.Given ${

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$4{x^2} + 4{y^2} + 30x - 13y - 25 = 0$

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(B) The family of circles passing through the intersection of two circles ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} + \lambda {S_2} = 0$.Given ${S_1} = {x^2} + {y^2} + 13x - 3y = 0$ and ${S_2} = {x^2} + {y^2} + 2x - 3.5y - 12.5 = 0$ (dividing the second equation by $2$).The equation is $({x^2} + {y^2} + 13x - 3y) + \lambda ({x^2} + {y^2} + 2x - 3.5y - 12.5) = 0$.$(1 + \lambda ){x^2} + (1 + \lambda ){y^2} + (13 + 2\lambda )x - (3 + 3.5\lambda )y - 12.5\lambda = 0$.Dividing by $(1 + \lambda )$,the centre is $\left( { - \frac{{13 + 2\lambda }}{{2(1 + \lambda )}}, \frac{{3 + 3.5\lambda }}{{2(1 + \lambda )}}} \right)$.Since the centre lies on $13x + 30y = 0$,we have $13\left( { - \frac{{13 + 2\lambda }}{{2(1 + \lambda )}}} \right) + 30\left( { \frac{{3 + 3.5\lambda }}{{2(1 + \lambda )}}} \right) = 0$.$-169 - 26\lambda + 90 + 105\lambda = 0 \Rightarrow 79\lambda = 79 \Rightarrow \lambda = 1$.Substituting $\lambda = 1$ into the family equation: $2{x^2} + 2{y^2} + 15x - 6.5y - 12.5 = 0$.Multiplying by $2$,we get $4{x^2} + 4{y^2} + 30x - 13y - 25 = 0$.

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

The equation of the circle which passes through the intersection of ${x^2} + {y^2} + 13x - 3y = 0$ and $2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and whose centre lies on $13x + 30y = 0$ is

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