If the circles x² + y² + 6x + 8y + 16 = 0 and ${\mathrm{x}}^2+{\mathrm{y}}^2+2(3-\sqrt{3})\mathrm{x}+\mathrm{x}+2(4-\sqrt{6})\mathrm{y}=\mathrm{k}+6\sqrt{3}+8\sqrt{6},\mathrm{k}>0$, touch internally at the point $\mathrm{P}(\mathrm{\alpha },\mathrm{\beta })$, then $(\mathrm{\alpha }+\sqrt{3}{)}^2+(\mathrm{\beta }+\sqrt{6}{)}^2$ is equal to __________

x² + y² + 6x + 8y + 16 = 0 has centre (-3,-4) and radius=3

$\begin{array}{l}{x}^2+y^2+2(3-\sqrt{3})x+2(4-\sqrt{6})y=k+6\sqrt{3}+8\sqrt{6},k>0\\ \text{ has centre }(\sqrt{3}-3,\sqrt{6}-4)\\ \text{ and radius }\sqrt{k+34}\end{array}$

These two circles touch internally

$$\begin{gathered} C_1{C}_2=\left|r_1-r_2\right| \\ \Rightarrow \sqrt{3+6}=\left|\sqrt{k+34}-3\right| \\ \therefore k=2\left(k>0\right) \\ \text{ Equation of common tangent to two circles is } \\ 2\sqrt{3}x+2\sqrt{6}y+16+6\sqrt{3}+8\sqrt{6}+k=0 \\ substitute k=2 \\ x+\sqrt{2}y+3+4\sqrt{2}+3\sqrt{3}=0 \\ \because \left(\alpha ,\beta \right)\text{ are foot of perpendicular from}\left(-3,-4\right)\text{ to line }\left(i\right)\text{ then } \\ \begin{array}{l}\frac{\alpha +3}{1}=\frac{\beta +4}{\sqrt{2}}=\frac{-(-3-4\sqrt{2}+3+4\sqrt{2}+3\sqrt{3})}{1+2}\\ \therefore \alpha +3=\frac{\beta +4}{\sqrt{2}}=-\sqrt{3}\\ \Rightarrow (\alpha +\sqrt{3}{)}^2=9\text{ and }(\beta +\sqrt{6}{)}^2=16\\ \therefore (\alpha +\sqrt{3}{)}^2+(\beta +\sqrt{6}{)}^2=25\end{array} \end{gathered}$$