If the circles x 2 + y 2 + 6 x + 8 y + 16 = 0 and x 2 + y 2 + 2 3 - 3 x + 2 4 - 6 y = k + 6 3 + 8 6 , k &#62…

The circle $x^2+y^2+6x+8y+16=0$ has centre $\left(-3,-4\right)$ and radius $\sqrt{9+16-16}=3$ units.

The circle $x^2+y^2+2\left(3-\sqrt{3}\right)x+2\left(4-\sqrt{6}\right)y=$ $k+6\sqrt{3}+8\sqrt{6},k>0$ has centre $\left(\sqrt{3}-3,\sqrt{6}-4\right)$ and radius $\sqrt{{\left(\sqrt{3}-3\right)}^2+{\left(\sqrt{6}-4\right)}^2+k+6\sqrt{3}+8\sqrt{6}}=\sqrt{k+34}$

Given that these two circles touch internally, so

distance between their centres$=\left|\text{difference of radii}\right|$

$\sqrt{3+6}=\left|\sqrt{k+34}-3\right|$

$\Rightarrow \sqrt{k+34}-3=\pm 3$

Here, $k=2$ is only possible value  $\left(\because k>0\right)$

Now the equation of common tangent to both the circles is given by $2\sqrt{3}x+2\sqrt{6}y+16+k+6\sqrt{3}+8\sqrt{6}=0$

$\because k=2$ then equation becomes 

$x+\sqrt{2}y+3\sqrt{3}+3+4\sqrt{2}=0 \cdots \left(\mathrm{i}\right)$

$\because \left(\alpha ,\beta \right)$ are foot of perpendicular from $\left(-3,-4\right)$ to this common tangent, then

$\frac{\alpha+3}{1}=\frac{\beta+4}{\sqrt{2}}=\frac{-(-3-4\sqrt{2}+3+4\sqrt{2}+3\sqrt{3})}{1+2}$ $\therefore \alpha+3=-\sqrt{3}$ and $\frac{\beta+4}{\sqrt{2}}=-\sqrt{3}$ $\Rightarrow (\alpha+\sqrt{3})^2=9$ and $(\beta+\sqrt{6})^2=16$ Hence, $(\alpha+\sqrt{3})^2+(\beta+\sqrt{6})^2=25$