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