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 radius squared is $g_1^2+f_1^2-c_1 = 10^2+0^2-4 = 96$.For two circles to cut orthogonally,the condition is $2g_1g_2 + 2f_1f_2 = c_1+c_2$.Here,$g_1 = -10, f_1 = 0, c_1 = 4$ and $g_2 = g, f_2 = f, c_2 = c$.So,$2(-10)(g) + 2(0)(f) = 4+c$,which gives $c = -20g-4$.The circle touches the line $x=2$,so the distance from the centre $(-g, -f)$ to $x=2$ is equal to the radius $r = \sqrt{g^2+f^2-c}$.$| -g - 2 | = \sqrt{g^2+f^2-c} \Rightarrow (g+2)^2 = g^2+f^2-c$.$g^2+4g+4 = g^2+f^2-c \Rightarrow f^2-c-4g-4 = 0$.Substituting $c = -20g-4$ into the equation:$f^2 - (-20g-4) - 4g - 4 = 0$ $\Rightarrow f^2 + 20g + 4 - 4g - 4 = 0$ $\Rightarrow f^2 + 16g = 0$.Replacing $(-g, -f)$ with $(x, y)$,we have $g = -x$ and $f = -y$.$(-y)^2 + 16(-x) = 0$ $\Rightarrow y^2 - 16x = 0$ $\Rightarrow y^2 = 16x$.

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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