Consider a coordinate system \vec{r'}=(x', y') with its origin at the centre of the circle. In this coordinate system the equation of the circle is x'^2 + y'^2 = a^2.
Suppose if the centre of the circle is located at the point \vec{r_0}=(x_0, y_0) in the original coordinate system.
A point on the circle has a coordinate \vec{r_{}}=(x, y) in the original coordinate and \vec{r'} = (x', y') in the circle centric coordinate. \vec{r_{}} and \vec{r'} are related by \vec{r'}=\vec{r_{}}-\vec{r_{0}}.
\vec{r'}=\vec{r_{}}-\vec{r_{0}}
(x', y') = (x, y) - (x_0, y_0) = (x-x_0, y-y_0)
x'^2+y'^2=a^2
(x-x_0)^2+(y-y_0)^2=a^2
Comparing this to the given equation (x+9)^2+(y-12)^2=62, we find that x_0 = -9: \quad y_0=+12 and a=\sqrt{64}.
So the position vector of the circle centre in the original coordinate system is \vec{r_0}=(-9, +12). It is at a distance of r_0 = \sqrt{(-9)^2+12^2}=15 units at an angle atan(\frac{12}{-9})=126.869876^{o} counter-clockwise to the X axis.
