What are the coordinates of the center of the circle given 3 points

Question

if we have been given the 3 points $A (ax, ay) , B (bx,by) \text{ and } C(cx, cy) $ that are not on a line.

What are the coordinates of the centre of the circle that goes trough $ A, B, \text{ and } C $ ?

Answer 1

We write the equation of lines $AB$ and $BC$. Find the midpoint of each line. Write the equation of perpendicular bisector of each line and solve them simultaneously. The crossing point is the coordinate of the circle.

Answer 2

$$ (x-h)^2+(y-k)^2 = r^2 $$ 3 equations, 3 unknown variables (h,k,r): $$ (A_{ix}-h)^2+(A_{iy}-k)^2 = r^2 $$ done.

Answer 3

If you are familiar with solving multiple simultaneous equations, then consider the following approach:

Since each of the points $A = (x_A, y_A)$, $B = (x_B, y_B)$, and $C = (x_C, y_C)$ are on the circumference of the circle, they are at distance $r$ from the center $(x, y)$. This gives us three simultaneous equations:

$$\begin{cases} (x_A - x)^2 + (y_A - y)^2 = r^2 \\ (x_B - x)^2 + (y_B - y)^2 = r^2 \\ (x_C - x)^2 + (y_B - y)^2 = r^2 \end{cases} \iff \begin{cases} x^2 - x_A x + y^2 - y_A y + x_A^2 + y_A^2 - r^2 &= 0 \\ x^2 - x_B x + y^2 - y_B y + x_B^2 + y_B^2 - r^2 &= 0 \\ x^2 - x_C x + y^2 - y_C y + x_C^2 + y_C^2 - r^2 &= 0 \end{cases}$$ Since there are three equations, you can use them to solve the three unknowns $x$, $y$, and $r$. (Although, you do not need $r$, and its expression is pretty nasty, so it's best to skip it.)

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I'm quite slow at solving multiple simultaneous equations, because I prefer to use tools like Maple to do the dull work for me.

So, if I needed to solve this without such tools, I'd first simplify the problem, by rotating, translating, and scaling the coordinates so that $A'=(0,0)$, $B'=(1,0)$, and $C'=(u,v)$ -- that is, $A$ is translated to origin, new $x$ axis is from $A$ towards $B$, and scale is such that the distance between $A$ and $B$ is one unit. The new $x$ axis vector then matches vector $\vec{e}_x$ in the old coordinate system, $$\vec{e}_x = B - A = (x_B - x_A, y_B - y_A)$$ The $y$ vector must be perpendicular to that, so we just rotate it 90 degrees counterclockwise: $$\vec{e}_y = (y_B - y_A, x_A - x_B)$$ $C'$ (point $C$ in the new coordinates) is then $$\begin{cases} u = \left(C - A\right) \cdot \frac{\vec{e}_x}{\lVert \vec{e}_x \rVert^2} = \frac{(x_C - x_A)(x_B - x_A) + (y_C - y_A)(y_B - y_A)}{(x_B - x_A)^2 + (y_B - y_A)^2} \\ v = \left(C - A\right) \cdot \frac{\vec{e}_y}{\lVert \vec{e}_y \rVert^2} = \frac{(x_C - x_A)(y_B - y_A) + (y_C - y_A)(x_A - x_B)}{(x_B - x_A)^2 + (y_B - y_A)^2 } \end{cases}$$

In these new coordinates we now have a much simpler triplet of equations: $$\begin{cases} {x'}^2 + {y'}^2 = {r'}^2 \\ (x'-1)^2 + {y'}^2 = {r'}^2 \\ (x'-u)^2 + (y'-v)^2 = {r'}^2 \end{cases} \iff \begin{cases} {x'}^2 + {y'}^2 - {r'}^2 = 0 \\ {x'}^2 + {y'}^2 - {r'}^2 - 2 {x'} + 1 = 0 \\ {x'}^2 + {y'}^2 - {r'}^2 - 2 u {x'} + u^2 - 2 v {y'} + v^2 = 0 \end{cases}$$ As you would guess, or can calculate by substracting the second equation from the first, we have $$2 x' - 1 = 0 \iff x' = \frac{1}{2}$$ which should be obvious from $A'=(0,0)$ and $B'=(1,0)$: all points at the same distance from both must have $x'=1/2$.

(Do note the similarity to the pure geometric solution: perpendicular bisector of the edge $AB$. We can rotate and scale any pair of the three points to $(0,0)$ and $(1,0)$, and in all cases we end up with $x' = 1/2$. This directly means that the intersection of the three edges' perpendicular bisectors (lines perpendicular to the edge, intersecting the edge at the midpoint of the edge) must be the center of the circle. Neat.)

We can then solve for $y'$ by substituting $x'=1/2$ into the third equation. It'll simplify to a quadratic equation, and we can definitely solve that analytically. We end up with $$y' = v \pm \sqrt{{r'}^2 - u^2 + u - 1/4}$$ Only one of those is at distance $r'$ from origin (and thus $A'$ and $B'$), so after checking which sign you need for that, you have $x'$ and $y'$.

To go back to original coordinates is easy: $$\vec{p} = (x, y) = A + \vec{e}_x x' + \vec{e}_y y'$$ or, equivalently, $$\begin{cases} x = x_A + x' (x_B - x_A) + y' (y_B - y_A) \\ y = y_A + x' (y_B - y_A) + y' (x_A - x_B) \end{cases}$$