Even the direction of the arc is defined(clockwise or counter-clockwise), there are generally two possible solutions, because the radius angle could be less or greater than $\pi$. Let me show you by the planar vector method:
Given the start point $A(x_0,y_0)$and end point $B(x_1,y_1)$, let $|\vec{AB}|=2l$, so $l=\frac{\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}}{2}$. Let the midpoint of $AB$ is point $M$, we know the center point $O$ is subject to $OM\perp AB$, and $|\vec{AM}|=l,|\vec{MO}|=\sqrt{r^2-l^2}$.
Now, $\vec{AM}=(\frac{x_1-x_0}{2}+i(\frac{y_1-y_0}{2}))$, since $OM\perp AB$, we get $\vec{MO}=\frac{|\vec{MO}|}{|\vec{AM}|}\cdot\vec{AM}\times (cos\theta+i\cdot sin\theta)$, where $\theta=\frac{\pi}{2}$ or $\theta=-\frac{\pi}{2}$, depending on the rotary direction of the arc and whether the radius angle is less than $\pi$.
Case 1: $\theta=\frac{\pi}{2}$. we get $\vec{MO}=\frac{\sqrt{r^2-l^2}}{l}\cdot[(-\frac{y_1-y_0}{2})+i(\frac{x_1-x_0}{2})]$. Since $\vec{MO}=(x-\frac{x_0+x_1}{2})+i(y-\frac{y_0+y_1}{2})$, comparing the real and imaginary part, we konw $x=\frac{x_0+x_1}{2}-\frac{(y_1-y_0)\sqrt{r^2-l^2}}{2l}$ and $y=\frac{y_0+y_1}{2}+\frac{(x_1-x_0)\sqrt{r^2-l^2}}{2l}$.
Case 2: $\theta=-\frac{\pi}{2}$.we get $\vec{MO}=\frac{\sqrt{r^2-l^2}}{l}\cdot[(\frac{y_1-y_0}{2})-i(\frac{x_1-x_0}{2})]$. Comparing again, we know $x=\frac{x_0+x_1}{2}+\frac{(y_1-y_0)\sqrt{r^2-l^2}}{2l}$ and $y=\frac{y_0+y_1}{2}-\frac{(x_1-x_0)\sqrt{r^2-l^2}}{2l}$.
When shold $\theta$ be $\frac{\pi}{2}$ or $-\frac{\pi}{2}$?
By drawing a circle and observing the different cases, it's easy to know: $counter-clockwise\ and\ radius\ angle\ is \gt\pi: \theta=-\frac{\pi}{2}$.
$counter-clockwise\ and\ radius\ angle\ is \lt\pi: \theta=\frac{\pi}{2}$.
$clockwise\ and\ radius\ angle\ is \gt\pi: \theta=\frac{\pi}{2}$.
$clockwise\ and\ radius\ angle\ is \lt\pi: \theta=-\frac{\pi}{2}$.
Also, you can see there is no solution when $r\lt l$ and there is only one solution when $l=r$.