Find  angle between the circles given by the equations ${\mathrm{x}}^2+{\mathrm{y}}^2-12\mathrm{x}-6\mathrm{y}+41=0,{\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}+6\mathrm{y}-59=0$

Given circles are
$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2-12\mathrm{x}-6\mathrm{y}+41=0,{\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}+6\mathrm{y}-59=0 \\ {\text{ centre C}}_1=(-\mathrm{g},-\mathrm{f}) {\text{ centre C}}_2=(-\mathrm{g},-\mathrm{f}) \\ = (6, 3) = (–2, –3) \end{gathered}$$
$$\begin{gathered} \text{ Radius }{r}_1=\sqrt{g^2+f^2-c} \text{ Radius }{r}_2=\sqrt{g^2+f^2-c} \\ \begin{array}{l}=\sqrt{36+9-41} =\sqrt{4+9+59}\\ =2 =\sqrt{72}=6\sqrt{2} \end{array} \end{gathered}$$
Distance between centres
$\mathrm{d}=C_1{C}_2=\sqrt{64+36}=10$.
$\text{'}\mathrm{\theta }\text{'}$ is angle between the circles.
Then $\cos \mathrm{\theta }=\frac{{\mathrm{d}}^2-{\mathrm{r}}_1^2-{\mathrm{r}}_2^2}{2{\mathrm{r}}_1{\mathrm{r}}_2}$
$\begin{array}{l}=\frac{{10}^2-2^2-(\sqrt{72}{)}^2}{2\cdot 2\cdot 6\sqrt{2}}\\ =\frac{100-4-72}{2\cdot 2\cdot 6\sqrt{2}}\\ =\frac{24}{24\sqrt{2}}\\ =\frac{1}{\sqrt{2}}\\ \Rightarrow \mathrm{\theta }={45}^0\text{ or }\frac{\mathrm{\pi }}{4}\end{array}$