Locus of the centre of circle of radius 2  which rolls on out side the rim of the circle

$x^2+y^2-4x-6y-12=0$ is

  The locus of the centres is again a circle (outermost) concentric with the given circle

                                        $C=\left(2,3\right)$

$r_1=\sqrt{4+9+12}=5$

$r_2=5+2=7$

$\therefore {\left(x-2\right)}^2+{\left(y-3\right)}^2=49$                  

$\Rightarrow x^2+y^2-4x-6y-36=0$