Thank you for reporting, we will resolve it shortly
Q.
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
TS EAMCET 2020
Solution:
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 $C_{2}: 2(g)(-5)+2(f)(6)=c+52$
$10 g-12 f=-c-52 \ldots$.(ii)
Subtracting on Eqs. (i) and (ii), we get
$-8 g+16 f=56$
or $g-2 f=-7$ or $g=2 f-7 \ldots$ (iii)
Multiplying Eq. (i) by 5 and then subtracting Eq. (ii), we get
or $8 f=-c+18$ 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 $f-2=0$
$\Rightarrow f=2$
Putting the value of ' $f$ in Bq. (iii), we get
$g=2 \times 2-7 \Rightarrow g=-3$
Therefore, coordinate of centre is $(+3,-2)$.
