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

<p>The circle ${x^2} + {y^2} + 6x + 8y + 16 = 0$ has centre $( - 3, - 4)$ and radius 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 $\sqrt {k + 34}$</p> <p>$\because$ These two circles touch internally hence</p> <p>$\sqrt {3 + 6} = \left| {\sqrt {k + 34} - 3} \right|$</p> <p>Here, $k = 2$ is only possible ($\because$ $k > 0$)</p> <p>Equation of common tangent to two circles is $2\sqrt 3 x + 2\sqrt 6 y + 16 + 6\sqrt 3 + 8\sqrt 6 + k = 0$</p> <p>$\because$ $k = 2$ then equation is</p> <p>$x + \sqrt 2 y + 3 + 4\sqrt 2 + 3\sqrt 3 = 0$ ...... (i)</p> <p>$\because$ ($\alpha$, $\beta$) are foot of perpendicular from $( - 3, - 4)$</p> <p>To line (i) then</p> <p>$${{\alpha + 3} \over 1} = {{\beta + 4} \over {\sqrt 2 }} = {{ - \left( { - 3 - 4\sqrt 2 + 3 + 4\sqrt 2 + 3\sqrt 3 } \right)} \over {1 + 2}}$$</p> <p>$\therefore$ $\alpha + 3 = {{\beta + 4} \over {\sqrt 2 }} = - \sqrt 3$</p> <p>$\Rightarrow {\left( {\alpha + \sqrt 3 } \right)^2} = 9$ and ${\left( {\beta + \sqrt 6 } \right)^2} = 16$</p> <p>$\therefore$ $${\left( {\alpha + \sqrt 3 } \right)^2} + {\left( {\beta + \sqrt 6 } \right)^2} = 25$$</p>