# Determine origin of a circle given ordered pairs.

Given ordered pairs $p_i = (x_i, y_i)$ where $x, y \in I$, find a pair $(x_o, y_o)$ where the distance between $o$ and all $p_i$ is equal.

The problem may also be imagined as trying to find the origin of a unit circle, given points that exist along its edge. I've been wracking my brain trying to figure out how.

I know that the slope may be calculated with two points as $m = \frac{y_2-y_1}{x_2-x_1}$ and further, the angle via $\theta=\arctan(m)$. I also know that where $\theta = 45^\circ, i = 2$, the origin of the circle is $(x_1, y_2),(x_2, y_1)$. I can imagine extrapolating the origin via $(x_2+((x_2-x_1)-(y_2-y_1)), y_1),x_2-x_1<y_2-y_1$. I don't think that's completely correct. The idea is to add to the coordinate with the smallest difference until the difference between the differences of both x and y is equal (create a 45 degree angle).

Anyways, thanks so much for any help. If you can, draw a picture because I find it somewhat difficult to visualize mathematics. If you don't have the chance to draw a picture, please provide a written explanation of anything you use more complex than algebra (I'm still on mathematical training wheels).

You can only guarantee to do this in two dimensions with three points, when you are effectively trying to find the circumcentre of a triangle. This is done by finding the perpendicular bisectors of the sides, and locating their point of intersection.

Equivalently if the points are $(x_1,y_1); (x_2,y_2); (x_3, y_3)$ then the centre of the circle satisfies the equations you get by equating the distances to the three points, so

$$(x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2$$ which becomes $$2(x_2-x_1)x+2(y_2-y_1)y=x_2^2-x_1^2+y_2^2-y_1^2$$

And the similar equation given by using the third point instead of the second one.

This gives you two (simultaneous linear) equations in the two unknowns $x,y$ which you can solve. I'll leave you to work out a nice symmetrical form for the answer ...

Let $P_1$, $P_2$ and $P_3$ be three points on a circle $C$. Let $L$ be the line equidistant from $P_1$ and $P_2$, i.e. the perpendicular bisector of the line segment from $P_1$ to $P_2$, and similarly let $M$ be the line equidistant from $P_1$ and $P_3$. Then $L$ and $M$ intersect in the centre of the circle $C$.

• This is the one which is most intuitive. It's definitely a fine enough explanation to receive an upvote, but it does gloss over the mathematics in favour of simplicity. – Aarowaim May 4 '14 at 18:47

The distance between $p_i$ and $p_o$ is $r=\sqrt{(x_i-x_o)^2+(y_i-y_o)^2}$

So take three of your points, $p_1, p_2, p_3$ and equate distances and solve for $p_o$.

$$(x_1-x_o)^2+(y_1-y_o)^2=r^2$$ $$(x_2-x_o)^2+(y_@-y_o)^2=r^2$$ $$(x_3-x_o)^2+(y_3-y_o)^2=r^2$$