<s>[INST] Find center of circle, which tangent to $y=2$ at $(3,2)$ and $y=-x\sqrt 3 +2$ Using $$r=\frac{|Ax + By + c|}{\sqrt{A^2+B^2}}$$
I get $$b=2+a\sqrt3$$ and stuck,because the options are on numbers. What do i do next?
 [/INST] Suppose $C=(a,b)$ is the center of the circle. 
Then $a=3$ because the circle is tangent to $y=3$ at point $(3,2)$.
Also the distance of $C$ from both lines $y=2$ and $y=-x\sqrt{3}+2$ is same:
$$|b-2|=\frac{|b+3\sqrt{3}-2|}{2}$$
and note that we have two solutions for $b<2$ and $b>2$. 
can you proceed?

Edit: If $b>2$, then $b+3\sqrt{3}-2>3\sqrt{3}>0$, thus we have
$$b-2=\frac{b+3\sqrt{3}-2}{2}$$
which gives $b=2+3\sqrt{3}$.
If $b<2$, $|b-2|=2-b$, but $|b+3\sqrt{3}-2|$ is $\pm(b+3\sqrt{3}-2)$
thus $$2-b=\frac{b+3\sqrt{3}-2}{2}$$
which gives $b=2-\sqrt{3}$ or
$$2-b=-\frac{b+3\sqrt{3}-2}{2}$$
which leads to repeated value $b=2+3\sqrt{3}$
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