Calculate circle given two points and conditions there are a lot of ways to calculate a circle given two points given the formula $(x-k)^2+(y-k)^2=r^2$ on this board. However these are not unique as there are arbitrary many possibilities to draw a circle through two points (with different radius).
I want to add another condition, such that there is only one possible solution. My condition is that one point is the highest point. To sketch my problem, I created an image with R.

As you can see in this specific image, the center is given for $p_0=(0,0)$ (blue dot, generally unknown), the highest point is given for $p_1=(x_1,y_1)$ (green dot, highest value, known) and another point $p_2=(x_2,y_2)$ (red dot, known value) is given as well. The red and green lines have the length of the desired radius $r$ (unknown) and the gray line is known through points $p_1$ and $p_2$.
So to sum up, I know the two values $p_1$ and $p_2$ and I'm searching for the radius $r$. I thought that the gray line might give me some information utilizing an equal-sided triangle, but so far I had no luck.
So my question is: How can I calculate radius $r$ given two points $p_1$ and $p_2$.
 A: The highest point is directly above the center.  The center has to be on the perpendicular bisector of the segment $p_1p_2$ (all points equidistant from $p_1, p_2$ are on the perpendicular bisector, and the center is exactly $r$ away from both).
So the center is the intersection of a vertical line through the highest point and the perpendicular bisector of the segment $p_1p_2$.  Then $r$ is easy to calculate.
A: The highest point lies directly above the center of the circle.  So if the center is an unknown $(x,y) $ then the highest point is $(x,y+r)=(x_1,y_1) =p_1$.  Or $x=x_1;y=y_1-r $.
So if $p_2=(x_2,y_2) $ then $r=\sqrt {(x_2-x_1)^2+(y_2-y_1+r)^2} $.
Solving we get: 
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====== old answer =======
If $(x,y)$ is the "highest" point then the tangent line to that circle is horizontal and the center of the circle is $(x,y-r) $ (ie the center of the circle lies "directly" below $(x,y) $ a distance of $r$ lower.)
So given the other point is $(u,v) $ we need to solve that $(u-x)^2+(v-y+r)^2= r^2$. (i.e the distance from $(u,v) $ to $(x,y-r) $ is $r$.
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So $(u-x)^2+(v-y)^2+2r (u-y)+r^2=r^2$
So $r=\frac { (u-x)^2+(v-y)^2 }{2 (y-u)}$
A: Let $t$ be the angle between the green and the grey line. The length of the grey line is $|p_1p_2|=2r\cos t$.
Using the given coordinates for $p_1$ and $ p_2$ we also have $|p_1p_2|\cos t=(y_1-y_2)$.
We can now solve for $r$ finding $$r={1\over 2}{ |p_1p_2|^2\over (y_1-y_2)}$$ and the coordinates for the centre are $$C=(x_1,y_1-r)$$
