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>0, touch internally at the point P(α, β), then (α+√ 3…

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