The radius of the circle whose center lies at (1,2) while cutting the circle  $x^2+y^2+4x+16y-30=0$  orthogonally is____ units

$$\begin{gathered} x^2+y^2+4x+16y-30=0----->Eq.\left(1\right) \\ {x}^2+y^2+2gx+2fy+c^1=0 \\ {x}^2+y^2-2x+4y+c^1=0----->Eq.\left(2\right) \\ Eq.\left(1\right)\perp Eq.\left(2\right) \\ 2g{y}^{|}+2f{f}^{|}=c+c^{|} 4(-1)+\left(16\right)(-2)=-30+c^{|} \\ c=-36+30=-6 \therefore x^2+y^2-2x-4y-6=0 \\ Radius r=\sqrt{1+4+6}=\sqrt{11} \end{gathered}$$