The centre of the smallest circle which cuts | vedclass

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(C) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$. The centre is $(-g,-f)$.Given circles are $C_1: x^2+y^2-2x-4y-4=0$ and $C_2: x^2

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$(3,-2)$

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(C) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$. The centre is $(-g,-f)$.Given circles are $C_1: x^2+y^2-2x-4y-4=0$ and $C_2: x^2+y^2-10x+12y+52=0$.For $C_1$,$g_1=-1, f_1=-2, c_1=-4$. For $C_2$,$g_2=-5, f_2=6, c_2=52$.The condition for two circles to cut orthogonally is $2g_1g_2+2f_1f_2=c_1+c_2$.For $C_1$: $2g(-1)+2f(-2)=c-4$ $\Rightarrow -2g-4f=c-4$ $\Rightarrow 2g+4f=-c+4$ $(i)$.For $C_2$: $2g(-5)+2f(6)=c+52$ $\Rightarrow -10g+12f=c+52$ $\Rightarrow 10g-12f=-c-52$ (ii).From $(i)$,$c = 4-2g-4f$. Substituting into (ii): $10g-12f = -(4-2g-4f)-52 = -4+2g+4f-52 = 2g+4f-56$.$8g-16f = -56$ $\Rightarrow g-2f = -7$ $\Rightarrow g = 2f-7$ (iii).Substitute $c = 4-2(2f-7)-4f = 4-4f+14-4f = 18-8f$.The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(2f-7)^2+f^2-(18-8f)} = \sqrt{4f^2-28f+49+f^2-18+8f} = \sqrt{5f^2-20f+31}$.To minimize $r^2 = 5(f^2-4f)+31 = 5(f-2)^2+11$,we set $f=2$.Then $g = 2(2)-7 = -3$.The centre is $(-g,-f) = (3,-2)$.

The centre of the smallest circle which cuts the circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2-10x+12y+52=0$ orthogonally is

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