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?


Tangent lines meet at $B=(0,2)$,
and the minor angle at $B$ is $60^\circ$,
so we can construct an equilateral $\triangle ABC$,
for which the minor and major circles tangent to $y=2$ at $T=(3,2)$
are the inscribed and escribed circles, respectively.
The side length of the equilateral $\triangle ABC$
$|BC|=|CA|=|AB|=a=b=c=|AB|=2\cdot|BT|=6$.
So, the radius of the inscribed circle is
\begin{align}
r&=\tfrac13\cdot a\cdot\tfrac{\sqrt3}2
=\sqrt3
.
\end{align} 
And the radius of the escribed circle is
\begin{align}
r_c&
=\frac{a+b+c}{a+b-c}\cdot r
=\frac{3\,a}{a}\cdot r
=3\,r
,
\end{align} 
so the coordinates of the centers are
\begin{align} 
I&=(3,2-\sqrt3)
\approx(3,0.268)
,\\
O_c&=(3,2+3\sqrt3)
\approx(3,7.196)
.
\end{align} 
