<s>[INST] maximum value of $c$ 
If the circles $x^2+y^2+(3+\sin \beta)x+(2\cos \alpha)y=0$ and $x^2+y^2+(2\cos \beta)x+2cy=0$ touches each other. Then maximum value of $c$ is

Center $\displaystyle C_{1}:\bigg(\frac{-3-\sin \beta}{2},\frac{-2\cos \alpha}{2}\bigg)$ and $\displaystyle r_{1}=\sqrt{\frac{(3+\sin \beta)^2+4\cos^2 \alpha}{4}}$
Center $\displaystyle C_{2}:\bigg(\frac{-2\cos \beta}{2},-\frac{2c}{2}\bigg)$ and $\displaystyle r_{2}=\sqrt{\frac{4\cos^2 \beta+4c^2}{4}}$
if circle touches each other, then distance between center of $2$ circle is sum of radius of circle 
$$C_{1}C_{2}=r_{1}+r_{2}$$
$$\sqrt{(3+\sin \beta-2\cos \beta)^2+(2\cos \alpha -2c)^2}=\sqrt{(3+\sin \beta)^2+4\cos^2 \alpha}+\sqrt{4\cos^2 \beta+4c^2}$$
How do i solve it Help me please
 [/INST] Observe that both circles pass through $(0,0)$, no matter what $\alpha, \beta, c$ are. 
So their centers and $(0, 0)$ are on the same line, this implies 
$$
\frac{2c}{2\cos\alpha}=\frac{2\cos\beta}{3+\sin\beta},
$$
that is, $c=\frac{2\cos\beta\cos\alpha}{3+\sin\beta}$; at maximum it must be  $\cos\alpha=1$ (actually at $\cos\alpha=-1$ there is another maximum but it gives the same value and we can ignore it), and, by a quick calculation using derivative in $\beta$ we see it must be $\sin\beta=-\frac 13$ and $\cos\beta=\frac{\sqrt 8}{3}$ at maximum. So maximum of $c$ is $\frac {\sqrt 2}{2}$. 
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