What is the equation for a circumcircular arc with 3 points?

Question

I know that the equation of a circle is $(x – h)^2$ + $(y – k)^2$ = $r^2$, but how do I find the equation of circumcircular arc given 3 points?

Answer 1

If you draw the line connecting two points, and you take the perpendicular bisector of this line, you get a diameter of your circle. So if you do this with two of the three pairs, you get two diameters. Their intersection is thus the center.

Proving this is a cool exercise in Euclidean geometry, and the result gives a clear algorithm. Find these two lines, compute their intersection, and then find the distance from that point to any of the three points to get the radius.

Answer 2

If you want an algebraic approach:

$$(x_1-h)^2 + (y_1-k)^2 = r^2 \tag{1}$$

$$(x_2-h)^2 + (y_2-k)^2 = r^2 \tag{2}$$

$$(x_3-h)^2 + (y_3-k)^2 = r^2 \tag{3}$$

Subtracting $(3)$ from $(1)$,

$$2h(x_1-x_3)+2k(y_1-y_3)=x_1^2-x_3^2+y_1^2-y_3^2$$

Similarly, subtracting $(2)$ from $(1)$,

$$2h(x_1-x_2)+2k(y_1-y_2)=x_1^2-x_2^2+y_1^2-y_2^2$$

Now you have two linear equations in two unknowns, you can solve for $h$ and $k$ and then recover $r$.

Answer 3

Given coordinates of the three points $A,B,C$ that form a valid triangle $ABC$ with side lengths $a,b,c$, the coordinates of the center of the circumscribed circle can be found as

\begin{align} O&= A\cdot \frac{a^2\,(b^2+c^2-a^2)}{((b+c)^2-a^2)(a^2-(b-c)^2)} \\ &+B\cdot \frac{b^2\,(a^2+c^2-b^2)}{((a+c)^2-b^2)(b^2-(a-c)^2)} \\ &+C\cdot \frac{c^2\,(b^2+a^2-c^2)}{((b+a)^2-c^2)(c^2-(b-a)^2)} , \end{align}

and the radius of the circle can than be found as

\begin{align} R&=|OA|=|OB|=|OC| , \end{align}

or using the known formula \begin{align} R&=\frac{abc}{4S} , \end{align}
where $S$ is the area of $\triangle ABC$.