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?
 A: 
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} 
A: You can think of the entire thing geometrically. If you had two lines that were tangent to a circle and you wanted to draw said circle, here would be the steps:


*

*Construct a perpendicular line from a point on one of the given lines.

*Construct an angle bisector of the given lines.

*The constructed lines meet at the centre of the circle.


So this case is no exception. The thing is, you have to be smart about it.
Consider $f(x)=1 \cdot x$. This line forms an angle of $45^\circ$ with any horizontal line you draw on the plane(why?) But $\tan{45^\circ}=1$! This means that the slope of a line is intrinsically connected with the tangent of the angle it forms with a horizontal. No, it is the tangent of that angle!
The line $y=x(-\sqrt{3})$ forms an angle of $-60^\circ$ or $120^\circ$ with the horizontal and thus with the other given line. Therefore so should it's parallel counterpart $y=x(-\sqrt{3})+2$ which we ignored for a while. Thus the angle bisector line has a slope of $\tan{-30^\circ}$. It should also pass through the common point $(0,2)$ so it's y-intercept is 2. 
Equation of line 1 $\implies y=-x\dfrac{\sqrt{3}}{3}+2$
The second one is simple. From the point $(3,2)$, the perpendicular line would be:
Line 2 $\implies x=3$
Where the two lines meet is where the x value on line 1 $=3$, so the y value $=\dfrac{-3\sqrt{3}}{3}+2=2-\sqrt{3}$
What's the centre of the circle?
$(3, 2-\sqrt{3})$
What's the radius?
$=2-(2-\sqrt{3})=\sqrt{3}$
Have fun with those.
Edit: I just realised you could have drawn another circle on the other side of the lines. But it's the co-ordinate given that determines which circle we want. But I guess for one with the same radius use $(-3,2)$ instead
A: 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 $b=2+3\sqrt{3}$
