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

Let the equation of the required circle be $x^2+y^2+2 g x+2 f y+c=0$ and centre is $(-g,-f)$ Given circles, $C_1: x^2+y^2-2 x-4 y-4=0$ Here, $g_1=-1, f_1=-2, c_1=-4$ $C_2: x^2+y^2-10 x+12 y+52=0$ Here, $g_2=-5, f_2=6, c_2=52$ Condition of two circles cuts orthogonally $2 g_1 g_2+2 f_1 f_2=c_1+c_2$ For $C_1: 2(g)(-1)+2(f)(-2)=C-4$ or $2 g+4 f=-c+4$ $\ldots$ (i) For $C_2: \quad 2(g)(-5)+2(f)(6)=c+52$ $10 g-12 f=-c-52$ $\ldots$ (ii) Subtracting on Eqs. (i) and (ii), we get $-8 g+16 f=56$ or $\quad g-2 f=-7$ or $g=2 f-7$ $\ldots$ (iii Multiplying Eq. (i) by 5 and then subtracting Eq. (ii), we get
or $\quad 8 f=-c+18$ or $c=-8 f+18$ Hence, radius $r=\sqrt{g^2+f^2-c}$ $=\sqrt{(2 f-7)^2+f^2(-8 f+18)}$ $=\sqrt{5 f^2-20 f+31}$ $=\sqrt{5\left(f^2-4 f\right)+31}=\sqrt{5(f-2)^2+11}$ For minimum, value $f-2=0$ $\Rightarrow \quad f=2$ Putting the value of ' $f$ ' in Eq. (iii), we get $g=2 \times 2-7 \Rightarrow g=-3$ Therefore, coordinate of centre is $(+3,-2)$.