Let the latus rectum of the parabola y^{2} = | vedclass

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(A) The parabola is $y^{2} = 4x$. Comparing with $y^{2} = 4ax$,we get $a = 1$.The length of the latus rectum is $4a = 4(1) = 4$.The common chord of th

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$8$

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(A) The parabola is $y^{2} = 4x$. Comparing with $y^{2} = 4ax$,we get $a = 1$.The length of the latus rectum is $4a = 4(1) = 4$.The common chord of the two circles $C_{1}$ and $C_{2}$ is the latus rectum of the parabola,so its length is $4$.Let $D$ be one endpoint of the latus rectum and $B$ be the midpoint of the common chord. Thus,$DB = \frac{4}{2} = 2$.Let $r$ be the radius of the circles,$r = 2\sqrt{5}$.In the right-angled triangle formed by the radius,the distance from the center to the chord,and half the chord length,let $x$ be the distance from the center of a circle to the common chord.$x^{2} + DB^{2} = r^{2}$$x^{2} + 2^{2} = (2\sqrt{5})^{2}$$x^{2} + 4 = 20$$x^{2} = 16 \implies x = 4$.The distance between the centers $C_{1}$ and $C_{2}$ is $x + x = 4 + 4 = 8$.

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

Let the latus rectum of the parabola $y^{2} = 4x$ be the common chord to the circles $C_{1}$ and $C_{2}$,each of them having radius $2\sqrt{5}$. Then,the distance between the centres of the circles $C_{1}$ and $C_{2}$ is

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