Locus of the center of circle of radius 2 which rolls on out side the rim of the circle x² + y² – 4x – 6y – 12 = 0 is

x² + y² – 4x – 6y – 12 = 0

$$\begin{gathered} x^2 + y^2 – 4x – 6y – 12 = 0 \\ C=\left(2,3\right) , r=5 \end{gathered}$$

Locus is a concentric circle of radius r + 2

$$\begin{gathered} C=\left(2,3\right) , r=7 \\ Eq. circle is \\ {\left(h-2\right)}^2+{\left(k-3\right)}^2=49 \\ {h}^2+k^2-4h-6k+13=49 \\ {h}^2+k^2-4h-6k-36=0 \end{gathered}$$