Solution
We are given the equation of the circle
[tex]6x+37=20y-x^2-y^2[/tex]We will first rewrite the equation
[tex]\begin{gathered} x^2+y^2+6x-20y+37=0 \\ \text{Collect like terms} \\ x^2+6x+y^2-20y+37=0 \\ (x^2+6x)+(y^2-20y)=-37 \end{gathered}[/tex]Now we complete to perfect square in the bracket
[tex]\begin{gathered} x^2+6x \\ Co\operatorname{erf}ficientOfx=6 \\ DivideBy2=\frac{6}{2}=3 \\ SquareIt=3^2=9 \\ So\text{ we have} \\ x^2+6x+9-9 \\ (x+3)^2-9 \end{gathered}[/tex]Following the same process, we have
[tex]\begin{gathered} y^2-20y \\ so\text{ we will have} \\ (y-10)^2-100 \end{gathered}[/tex]Now, from
[tex]\begin{gathered} (x^2+6x)+(y^2-20y)=-37 \\ (x+3)^2-9+(y-10)^2-100=-37 \\ (x+3)^2+(y-10)^2=-37+9+100 \\ (x+3)^2+(y-10)^2=72 \\ \end{gathered}[/tex]Comparing with the equation of a circle
[tex](x-a)^2+(y-b)^2=r^2[/tex]Therefore,
[tex]\begin{gathered} Centre=(-3,10) \\ Radius^2=72 \\ Radius=\sqrt[]{72} \\ Radius=\sqrt[]{36\times2} \\ Radius=6\sqrt[]{2} \end{gathered}[/tex]