If the circles x ^ 2 + y ^ 2 + 6x + 8y + 16 = 0 and x ^ 2 + y ^ 2 +...

The circle ${x}^{2} + {y}^{2} + 6x + 8y + 16 = 0$ has centre \left(- 3, - 4\right)$and radius$\sqrt{9 + 16 - 16} = 3$units.</p><p>The circle${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}$</p><p>Given that these two circles touch internally, so</p><p>distance between their centres$= \left|\text{difference$of$radii}\right|$</p><p>$\sqrt{3 + 6} = \left|\sqrt{k + 34} - 3\right|

$⇒ \sqrt{k + 34} - 3 = \pm 3$

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

Now$the equation of common tangent to both the circles is given by2\sqrt{3}x + 2\sqrt{6}y + 16 + k + 6\sqrt{3} + 8\sqrt{6} = 0$

$∵ k = 2$ then equation becomes

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

$∵ \left(\alpha, \beta\right) are foot$ of $perpendicular from \left(- 3, - 4\right) to this common tangent, then</p><p>\dfrac{\alpha + 3}{1} =\dfrac{\beta + 4}{\sqrt{2}} = \dfrac{- \left(- 3 - 4\sqrt{2} + 3 + 4\sqrt{2} + 3\sqrt{3}\right)}{1 + 2}$

$∴ \alpha + 3 = - \sqrt{3} \text{\text{ and } } \dfrac{\beta + 4}{\sqrt{2}} = - \sqrt{3}$

$⇒ {\left(\alpha + \sqrt{3}\right)}$^{2} = 9 and ${\left(\beta + \sqrt{6}\right)}^{2} = 16$

Hence, ${\left(\alpha + \sqrt{3}\right)}$^{2} + {\left(\beta + \sqrt{6}\right)}^{2} = 25