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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?

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 $(1)$,

$$2h(x_1-x_3)+2k(y_1-y_3)=0$$

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

$$2h(x_1-x_2)+2k(y_1-y_2)=0$$

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

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.

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&=
Acdot frac{a^2,(b^2+c^2-a^2)}{((b+c)^2-a^2)(a^2-(b-c)^2)}
\
&+Bcdot frac{b^2,(a^2+c^2-b^2)}{((a+c)^2-b^2)(b^2-(a-c)^2)}
\
&+Ccdot 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 $triangle ABC$.