Finding the the radius of an arc between two lines. Essentially, I'm trying to recreate the 'Mini' logo in a program.
I know that one line extends from the centre 50.  There is a line 25 below it,
which extends 25 from the centre.  I'm trying to find the radius of the arc that connects them, and it's distance vertically from the centre point of the drawing.
From the diagram, I'm looking for the radius, l, and the distance for the centre, l.

 A: The center point of a circular arc going through $A$ and $B$ will sit somewhere on a line $L$ that (1) bisects the line drawn from $A$ to $B$, which we'll call $\overline{AB}$, and (2) is perpendicular to $\overline{AB}$.
I'm assuming the center point of the drawing is, vertically, halfway between $A$ and $B$, and we'll make it our origin.  Then we might assign $A$ the coordinates $(50, 12.5)$ and $B$ the coordinates $(25, -12.5)$.  We see that their vertical separation is indeed $25$, because $12.5-(-12.5)=25$.
(1) $L$ bisecting $\overline{AB}$ means that $L$ passes through the midpoint of $\overline{AB}$.  We can easily find the midpoint of $\overline{AB}$ by averaging the horizontal components and vertical components of $A$ and $B$.  The midpoint is then $(\frac{50+25}{2}, \frac{12.5-12.5}{2}) = (37.5, 0)$.
(2) To make $L$ perpendicular to $\overline{AB}$ we first have to know the slope of $\overline{AB}$.  The slope formula is $$m = \frac{y_b - y_a}{x_b - x_a} = \frac{12.5-(-12.5)}{50-25} = \frac{25}{25} = 1.$$ It is a neat property that, if $\overline{AB}$ has a slope $m$, every line perpendicular to $\overline{AB}$ will have a slope of $-\frac{1}{m}$.  In this case, that's $-1$.
We now have everything we need to determine an equation for $L$.  We'll write it in the form $y=nx+c$, where $n$ is the slope and $c$ is some constant.  We know one $x,y$ pair—the midpoint—and we also know the slope, -1.  So we just need to find $c$.  $$0 = -1\times 37.5 + c$$ $$0 = -37.5 + c$$ $$37.5 = c$$ And now we have a nice equation for $L$: $y = -x + 37.5$
If the center of our arc is above the center of the drawing, we know its $x$ coordinate is 0.  Since it's on the line $L$, it obeys the equation we just wrote, and we can use it to find the $y$ coordinate. $$y = -0 + 37.5 = 37.5$$  This is the "length from centre".
What is the radius?  It is the distance from $(0, 37.5)$ to $A$ or $B$.  We'll pick $A$, which we said was at $(50, 12.5)$. So we use the distance formula: $$d=\sqrt{(50-0)^2 + (12.5-37.5)^2}=\sqrt{50^2 + (-25)^2}=\sqrt{2500+625}=\sqrt{3125}\approx 55.9$$  To double check our work, we'll try $B$ as well.  $$d=\sqrt{(25-0)^2 + (-12.5-37.5)^2}=\sqrt{25^2 + (-50)^2}=\sqrt{625+2500}=\sqrt{3125}\approx 55.9$$  This is "radius of arc".
