The radius of the circle whose centre lies at ( 1 , 2 ) , while cutting the circle x 2 + y 2 + 4 x + 16 y -…

Given circle equation is $x^2+y^2+4x+16y-30=0$

By comparing with $x^2+y^2+2gx+2fy+c=0$

$2g=4, 2f= 16 \& c=-30$

Centre $= C_1 = \left(-g, -f\right) = \left(-2, -8\right)$

Radius $= r_1 = \sqrt{g^2+f^2-c} = \sqrt{4+64+30}= \sqrt{98}$

Other circle centre $C_2 = \left(1, 2\right)$

Distance between centres $= d =C_1{C}_2 = \sqrt{{\left(-3\right)}^2+{\left(10\right)}^2}=\sqrt{109}$

We know that when two circles are orthogonal $\Rightarrow d^2= {\left(r_1\right)}^2+ {\left(r_2\right)}^2$

$\Rightarrow 109=98+{r_2}^2$

$\therefore r_2 = \sqrt{11}$.