The locus of the centre of the circle,which c | vedclass

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(C) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$,where the centre is $(-g, -f)$.The centre of the given circle $x^2+y^2-20x+4=0$ is $(10, 0

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$y^2=16x$

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(C) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$,where the centre is $(-g, -f)$.The centre of the given circle $x^2+y^2-20x+4=0$ is $(10, 0)$ and its constant term is $4$.For two circles to cut orthogonally,the condition is $2(g_1g_2 + f_1f_2) = c_1 + c_2$.Substituting the values: $2(-g 	imes 10 + (-f) 	imes 0) = c + 4$,which gives $-20g = c + 4$,or $c = -20g - 4$.Since the circle touches the line $x=2$,the perpendicular distance from the centre $(-g, -f)$ to the line $x=2$ is equal to the radius $r = \sqrt{g^2+f^2-c}$.Thus,$|-g-2| = \sqrt{g^2+f^2-c}$.Squaring both sides: $(g+2)^2 = g^2+f^2-c$.$g^2+4g+4 = g^2+f^2-c$.$f^2 - 4g - c - 4 = 0$.Substituting $c = -20g - 4$: $f^2 - 4g - (-20g - 4) - 4 = 0$.$f^2 - 4g + 20g + 4 - 4 = 0$.$f^2 + 16g = 0$.Replacing $(-g, -f)$ with $(x, y)$,we have $g = -x$ and $f = -y$.$(-y)^2 + 16(-x) = 0$ $\Rightarrow y^2 - 16x = 0$ $\Rightarrow y^2 = 16x$.

List-$I$	List-$II$
$(A)$ Benzene	$1$. Phosgene
$(B)$ Ethylene	$2$. Silver mirror
$(C)$ Acetaldehyde	$3$. Mustard gas
$(D)$ Chloroform	$4$. $(4n + 2) \pi$ electrons
	$5$. Carbylamine

The locus of the centre of the circle,which cuts the circle $x^2+y^2-20x+4=0$ orthogonally and touches the line $x=2$,is

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Match the following lists:
List-$I$ List-$II$
$(A)$ Benzene $1$. Phosgene
$(B)$ Ethylene $2$. Silver mirror
$(C)$ Acetaldehyde $3$. Mustard gas
$(D)$ Chloroform $4$. $(4n + 2) \pi$ electrons
$5$. Carbylamine

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