Question

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

Answer 1

$ C _1(-3,-4) $
$ r _1=\sqrt{25-16}=3 $
$ C _2=(-3+\sqrt{3},-4+\sqrt{6}) $
$ r _2=\sqrt{34+ k } $
$ C _1 C _2=\left| r _1- r _2\right|$
$ C _1 C _2=\sqrt{3+6}=3$
$ 3=|3-\sqrt{34+ k }| \Rightarrow k =2 $
$ r _2=6$

$ (\alpha, \beta)=(-\sqrt{3}-3,-4-\sqrt{6}) $
$ (\alpha+\sqrt{3})^2+(\beta+\sqrt{6})^2=9+16=25$