The locus of the centre of a circle touching the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{y}-2\mathrm{x}=2\sqrt{3}-1$internally and tangents on which from the point (1, 2) are making an angle of 60º with each other is

$\mathrm{Equation} \mathrm{of} \mathrm{given} \mathrm{circle} \mathrm{is} {\left(\mathrm{x}-1\right)}^2+{\left(\mathrm{y}-2\right)}^2={\left(\sqrt{3}+1\right)}^2$

$\mathrm{Let} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{touching} \mathrm{given} \mathrm{circle} \mathrm{be} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right) \mathrm{and} \mathrm{radius} \mathrm{be} \mathrm{r}$

$\because \text{ circles touch internally }$

$\sqrt{{\left({\mathrm{x}}_1-1\right)}^2+{\left({\mathrm{y}}_1-2\right)}^2}=\sqrt{3}+1-\mathrm{r} \to$

$\sin 30°=\frac{r}{\sqrt{{\left(x_1-1\right)}^2+{\left(y_1-2\right)}^2}}$

$\sqrt{3}+1=3\mathrm{r}\Rightarrow \Rightarrow \mathrm{r}=\frac{\sqrt{3}+1}{3}$

$$\begin{gathered} {\left(x_1-1\right)}^2+{\left(y_1-2\right)}^2=\left({\left(\sqrt{3}+1-\left(\frac{\sqrt{3}+1}{3}\right)\right)}^2\right. \\ {\left({\mathrm{x}}_1-1\right)}^2+{\left({\mathrm{y}}_1-2\right)}^2=\frac{4}{9}(4+2\sqrt{3}) \end{gathered}$$