# What is the radius of this circle?

Using coordinates . . .

For convenience of notation, let $h=\sqrt{3}$.

Let $B = (-1,0),\;C=(1,0),\;A=(0,h)$.

Let $P$ be the center of the required circle.

Then $P=(0,1+r)$, where $r$ is the unknown radius.

Let $c(P,r)$ denote the circle centered at $P$, with radius $r$.

Let $M$ be the midpoint of segment $CA$.

Then $M=\bigl({\large{\frac{1}{2}}},{\large{\frac{h}{2}}}\bigr)$.

Let $s(M,1)$ denote the semicircle centered at $M$, with radius $1$, as shown in the diagram.

Let $T$ be the point where $c(P,r)$ meets $s(M,1)$.

Since $c(P,r)$ and $s(M,1)$ are tangent to each other at $T$, it follows that the points $M,P,T$ are collinear.

Then since $MT=1$ and $PT=r$, we get $MP=1-r$, hence by the distance formula \begin{align*} &MP^2=(1-r)^2\\[4pt] \implies\;&\left({\small{\frac{1}{2}}}-0\right)^{\!2}+\left({\small{\frac{h}{2}}}-(1+r)\right)^{\!\!2}=(1-r)^2\\[4pt] \implies\;&r=\frac{4h-1-h^2}{4(4-h)}\\[4pt] &\phantom{r}=\frac{4\sqrt{3}-4}{4(4-\sqrt{3})}\\[4pt] &\phantom{r}=\frac{3\sqrt{3}-1}{13}\approx .3227809558\\[4pt] \end{align*}

From the red triangle in the picture we infer $$(1-r)^2=\left({1\over2}\right)^2+\left(r+\left(1-{\sqrt{3}\over2}\right)\right)^2\ ,$$ and this leads to $$r={3\sqrt{3}-1\over13}\ .$$