If the circles x^{2}+y^{2}+6x+8y+16=0 and x^{ | vedclass

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(C) The circle $C_1: x^{2}+y^{2}+6x+8y+16=0$ has center $O_1(-3, -4)$ and radius $r_1 = \sqrt{3^2+4^2-16} = 3$.The circle $C_2: x^{2}+y^{2}+2(3-\sqrt{

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$25$

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(C) The circle $C_1: x^{2}+y^{2}+6x+8y+16=0$ has center $O_1(-3, -4)$ and radius $r_1 = \sqrt{3^2+4^2-16} = 3$.The circle $C_2: x^{2}+y^{2}+2(3-\sqrt{3})x+2(4-\sqrt{6})y = k+6\sqrt{3}+8\sqrt{6}$ has center $O_2(\sqrt{3}-3, \sqrt{6}-4)$ and radius $r_2 = \sqrt{(\sqrt{3}-3)^2 + (\sqrt{6}-4)^2 + k + 6\sqrt{3} + 8\sqrt{6}} = \sqrt{3-6\sqrt{3}+9 + 6-8\sqrt{6}+16 + k + 6\sqrt{3} + 8\sqrt{6}} = \sqrt{k+34}$.Since the circles touch internally,the distance between centers $d = |r_2 - r_1|$.$d^2 = ((\sqrt{3}-3) - (-3))^2 + ((\sqrt{6}-4) - (-4))^2 = (\sqrt{3})^2 + (\sqrt{6})^2 = 3+6 = 9$.Thus,$d = 3$.So,$3 = |\sqrt{k+34} - 3|$,which implies $\sqrt{k+34} = 6$ (since $k>0$),so $k+34 = 36$,$k=2$.The point of contact $P(\alpha, \beta)$ divides the line segment $O_1O_2$ externally in the ratio $r_1 : r_2 = 3 : 6 = 1 : 2$.Using the section formula for external division:$\alpha = \frac{r_1 x_2 - r_2 x_1}{r_1 - r_2} = \frac{1(\sqrt{3}-3) - 2(-3)}{1-2} = \frac{\sqrt{3}-3+6}{-1} = -\sqrt{3}-3$.$\beta = \frac{r_1 y_2 - r_2 y_1}{r_1 - r_2} = \frac{1(\sqrt{6}-4) - 2(-4)}{1-2} = \frac{\sqrt{6}-4+8}{-1} = -\sqrt{6}-4$.Therefore,$(\alpha+\sqrt{3})^{2} + (\beta+\sqrt{6})^{2} = (-3)^2 + (-4)^2 = 9 + 16 = 25$.

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

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