Is there a formula for finding the center point or radius of a circle given that you know two points on the circle and one of the points is perpendicular to the center?

I know that only having two points is not enough for determining the circle, but given that the center is on the same x coordinate as one of the points, is there a way to use those two points to find the center/radius of the circle? So you have the following data: x0 = 0 y0 = 0 x1 = 3 y1 = 1 y2 = ?

I want to build some ramps for my rc car and am trying to figure out the optimal curve for the ramps. The two points are the corners of a 3'x1' piece of plywood. I want to cut the best curve out of the plywood for the jump, and would like to have a formula to calculate/draw the curve for other size ramps.

Here is a diagram of the problem I am trying to solve. The rectangle will basically be a piece of plywood and the curve will be cut out of it. I am trying to solve for y2.

Ramp Circle

Basically, I am going to pin a piece of string in the ground y2 feet away from my board and attach a pencil to one end in order to mark the curve that I need to cut.

Thank you very much for your help.

  • 1
    $\begingroup$ You should say that the two points have the same x-coordinate, not that the points "are perpendicular". $\endgroup$ – Omnomnomnom Jan 2 '15 at 20:42

Let $A(0, 0), B(3, 1), M(0, r)$ (we place the point $A(x_0, y_0)$ on the origin)

Then the distance between A and M (d(A, M)) is r. The distance between B and M is also r, since A and B are both points on the circle.

$d(B, M)=\sqrt{(3-0)^2+(1-r)^2}=\sqrt{r^2-2r+10}=r$ (pythagorean theorem)

$r^2=r^2-2r+10$ (square both sides)

$r=5$ (solve for $r$)

Therefore, the coordinate of the middle point is 5 foot above the point $(x_0, y_0)$ and the radius is 5.

  • $\begingroup$ Super simple and it works. Thank you very much. I didn't even think about the distance formula. This makes me want to go back and practice the basics again. $\endgroup$ – BJ Anderson Jan 3 '15 at 2:13

Based on the diagram, we can solve the question as follows:

Because $C = (x_0,y_2)$ is equidistant from $P_0 = (x_0,y_0)$ and $P_1 = (x_1,y_1)$, $C$ must lie on the perpendicular bisector of $P_0$ and $P_1$.

The perpendicular bisector of two points is the line perpendicular to the line connecting them through their midpoint.

We calculate the midpoint $P$ as $$ P = \frac{P_0 + P_1}{2} = \left(\frac{x_0 + x_1}{2},\frac{y_0 + y_1}{2} \right) = (x_p,y_p) $$ The slope of the line connecting two points is given by the rise-over-run formula, and the perpendicular slope is its negative reciprocal. So, we have $$ m = - \frac{1}{\frac{y_1 - y_0}{x_1 - x_0}} = - \frac{x_1 - x_0}{y_1 - y_0} $$ Finally, the equation of a line through point $P$ and slope $m$ is given by the point slope formula. So, the perpendicular bisector is given by the equation $$ y - y_p = m(x - x_p) $$ $(x_0,y_2)$ lies on this line, so that $$ y_2 - y_p = m(x_0 - x_p) $$ Solving for $y_2$, we have $$ y_2 = m(x_0 - x_p) + y_p $$ all together, we have $$ y_2 = - \frac{x_1 - x_0}{y_1 - y_0}\left(x_0 - \frac{x_0 + x_1}{2}\right) + \frac{y_0 + y_1}{2} \implies\\ y_2 = - \frac{x_1 - x_0}{y_1 - y_0}\left(\frac{x_0 - x_1}{2}\right) + \frac{y_0 + y_1}{2} \implies\\ y_2 = \frac{(x_1 - x_0)^2}{2(y_1 - y_0)} + \frac{y_0 + y_1}{2} $$


It would help to convert this to a question about triangles instead. Pictured again below with a few modifications. (I'll use degrees as it is more common for household projects, but can easily be changed into radians as needed)

enter image description here

As the angle pointed to by the yellow arrow is $\arctan(\frac{1}{3})\approx 18.43^\circ$, that means the red angles are $90^\circ - \arctan(\frac{1}{3})\approx 71.57^\circ$

So, we have a $71.57, 71.57, 36.86$ triangle.

By the law of sines, $\frac{A}{\sin(a)}=\frac{B}{\sin(b)}$ you have $B = (\sqrt{3^2+1^2}\frac{\sin(71.57^\circ)}{\sin(36.86^\circ)}) \approx 5.0013$

  • $\begingroup$ You may want to use $\approx$ signs as the radius is actually 5. $\endgroup$ – Cyclohexanol. Jan 2 '15 at 21:00
  • $\begingroup$ indeed. Such is the trouble of taking only 4 sig figs on the angle measurements. $\endgroup$ – JMoravitz Jan 2 '15 at 21:02
  • $\begingroup$ To be more precise, with your method, the answer is $$\frac{\sqrt{(y_1-y_0)^2+(x_1-x_0)^2}*\sin(\frac{\pi}{2}-\tan^{-1}\left(\frac{|y1-y0|}{|x_1-x_0|}\right)}{\sin\left(\pi-2\left(\frac{\pi}{2}-\tan^{-1}\left({|y1-y0|}\over{|x_1-x_0|}\right)\right)\right)}$$ $\endgroup$ – Cyclohexanol. Jan 2 '15 at 21:10
  • $\begingroup$ Thank you (and everyone else) for your efforts. This is a nice, elegant solution and I would accept it if I could accept two answers. $\endgroup$ – BJ Anderson Jan 3 '15 at 2:15

So we have a circle through the origin and $(x,y)$ whose center lies in $(0,y_0)$.

By the pythagorean theorem, $$ y_0^2 = x^2+(y-y_0)^2 $$ so $x^2+y^2=2yy_0$ gives: $$ y_0 = \frac{x^2+y^2}{2y}.$$

  • $\begingroup$ This is close, but you left out a term. This should actually be x^2 + y^2 / 2y $\endgroup$ – BJ Anderson Jan 3 '15 at 2:59
  • $\begingroup$ @bj_anderson: thanks, fixed. $\endgroup$ – Jack D'Aurizio Jan 3 '15 at 11:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.