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

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Question

The circle $x^{2}+y^{2}+6 x+8 y+16=0$ has centre $(-3,-4)$ and radius 3 units.

The circle $x^{2}+y^{2}+2(3-\sqrt{3}) x+2(4-\sqrt{6}) y=$ $k+6 \sqrt

Vedclass · Accepted Answer

$25$

Vedclass · Answer

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

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 $\dots\dots$

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