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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).

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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 ...

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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 $C$.

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  • $\begingroup$ 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. $\endgroup$ – Aarowaim May 4 '14 at 18:47
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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 $p_o$.

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

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