How do I work out the point on a line at which an arc with a fixed start and end point is tangential to that point?

Question

Thank you for taking the time to read my question.

It's been an age since I studied mathematics and so my geometry is rusty ...

Please see the attached diagram - I am trying to connect points A and B with an arc that is tangential to a point C on the line DE. How do I find that point C?

I am using a CAD program to draw this but can't find the correct way of drawing the arc.

Any help would be appreciated! Thank you.Please see diagram of points listed above here.

Answer 1

The center will be on the perpendicular bisector to $AB$. We have $A(0,2500)$, $B(4775,2000)$, midpoint $M(2387.5,2250)$. The equation of the line $AB$: $y=\frac{2000-2500}{4775-0}x+2500=-\frac{20}{191}x+2500$, the equation of its perpendicular bisector: $y=\frac{191}{20}x+b$, plugging $M$ will give us $2250=\frac{191}{20} \cdot 2387.5+b \implies b=-20550.625$
The center will be $O(x, 9.55x-20550.625)$, equating distances will give us $$OA^2=x^2+(9.55x-20550.625-2500)^2=OC^2=(9.55x-20550.625-3000)^2$$ Solving this yields $x \approx 2014.79, y \approx -1309.38, r\approx 4309.38$.