Coordinates of circumcentre of a triangle in terms of triangle point coordinates

Question

For a triangle $ABC$, its circumcentre is the intersection of perpendicular bisectors of its three sides. Now for a particular triangle I can draw the bisectors and find the circumcentre. However I need a general formula to solve the position $(x,y)$ of the circumcentre, in terms of the positions of each point $A(x,y)$, $B(x,y)$ and $C(x,y)$ of the triangle. Is there such formula?

Answer 1

There is some point $(x,y)$

Such that $(x,y)$ is equidistant from A, B, C.

If $A = (x_a, y_a), B = (x_b, y_b), C = (x_c, y_c)$

$(x-x_a)^2 + (y-y_a)^2 = (x-x_b)^2 + (y-y_b)^2 = (x-x_c)^2 + (y-y_c)^2$

$x^2 + y^2 - 2xx_a - 2yy_a +x_a^2 + y_a^2 = x^2 + y^2 - 2xx_b - 2yy_b +x_b^2 + y_b^2 = x^2 + y^2 - 2xx_c - 2yy_c +x_c^2 + y_c^2$

We can subtract $x^2 + y^2$ from all 3 parts.

$2xx_a + 2yy_a + D = x_a^2+y_a^2\\ 2xx_b + 2yy_b + D = x_b^2+y_b^2\\ 2xx_c + 2yy_c + D= x_c^2+y_c^2$

And this is a system of linear equations with 3 unknowns.

I suppose we can apply Cramer's rule for a single formula.

$x = \frac {(x_a^2 + y_a^2)(y_b-y_c) + (x_b^2 + y_b^2)(y_c-y_a) + (x_c^2 + y_c^2)(y_a-y_b)}{2(x_a(y_b - y_c)+ x_b(y_c - y_a) + x_c(y_a-y_b))}$

$y = \frac {(x_a^2 + y_a^2)(x_b-x_c) + (x_b^2 + y_b^2)(x_c-x_a) + (x_c^2 + y_c^2)(x_a-x_b)}{2(y_a(x_b - x_c)+ y_b(x_c - x_a) + y_c(x_a-x_b))}$

Answer 2

In case there's not, you can easily derive it.
Perpendicular bisector to say $AB$ is all the points $X(x,y)$ that $|XA|=|XB|$: $$(x-x_A)^2+(y-y_A)^2=(x-x_B)^2+(y-y_B)^2$$ $$y (2 y_B - 2 y_A) = -x_A^2 - y_A^2 + x_B^2 + y_B^2 + (2 x_A - 2 x_B) x$$ Once you have perpendicular bisectors of say $AB$ and $AC$ you can compute their intersection -- the desired point.

Answer 3

Incredibly, I could not find the answer online, so let me try to derive it here.

The midpoint of $AB$ is $(\frac{x_a + x_b}{2}, \frac{y_a + y_b}{2})$. The slope of $AB$ is $\frac{y_b - y_a}{x_b - x_a}$. The slope perpendicular to this is the opposite of the inverse, namely $-\frac{x_b - x_a}{y_b - y_a}$. Thus, we can write the equation of the perpendicular bisector of $AB$ in point-slope form as $$ y - \frac{y_a + y_b}{2} = -\frac{x_b - x_a}{y_b - y_a} \left( x - \frac{x_a + x_b}{2} \right) $$ (Assuming $y_b - y_a \neq 0$). Applying the same reasoning to $AC$, the perpendicular bisector of that segment is $$ y - \frac{y_a + y_c}{2} = -\frac{x_c - x_a}{y_c - y_a} \left( x - \frac{x_a + x_c}{2} \right) $$ To find the intersection, we must find a solution for these two equations. Isolating $y$ on the left of both, we get $$ y = \frac{y_a + y_b}{2} -\frac{x_b - x_a}{y_b - y_a} \left( x - \frac{x_a + x_b}{2} \right) $$ $$ y =\frac{y_a + y_c}{2} -\frac{x_c - x_a}{y_c - y_a} \left( x - \frac{x_a + x_c}{2} \right) $$ so $$ \frac{y_a + y_b}{2} -\frac{x_b - x_a}{y_b - y_a} \left( x - \frac{x_a + x_b}{2} \right) =\frac{y_a + y_c}{2} -\frac{x_c - x_a}{y_c - y_a} \left( x - \frac{x_a + x_c}{2} \right) $$ $$ \frac{y_a + y_b}{2}- \frac{y_a + y_c}{2} + \left(-\frac{x_b - x_a}{y_b - y_a} x +\frac{x_b - x_a}{y_b - y_a} \frac{x_a + x_b}{2} \right) = \left( -\frac{x_c - x_a}{y_c - y_a}x +\frac{x_c - x_a}{y_c - y_a}\frac{x_a + x_c}{2} \right) $$ $$ \frac{y_a + y_b}{2}- \frac{y_a + y_c}{2} +\frac{x_b - x_a}{y_b - y_a} \frac{x_a + x_b}{2} - \frac{x_c - x_a}{y_c - y_a}\frac{x_a + x_c}{2} = \frac{x_b - x_a}{y_b - y_a} x -\frac{x_c - x_a}{y_c - y_a}x $$ $$ x= \frac{\frac{y_a + y_b}{2}- \frac{y_a + y_c}{2} +\frac{x_b - x_a}{y_b - y_a} \frac{x_a + x_b}{2} - \frac{x_c - x_a}{y_c - y_a}\frac{x_a + x_c}{2}}{\frac{x_b - x_a}{y_b - y_a} -\frac{x_c - x_a}{y_c - y_a}} $$

And applying the same formulae with x and y reversed

$$ y= \frac{\frac{x_a + x_b}{2}- \frac{x_a + x_c}{2} +\frac{y_b - y_a}{x_b - x_a} \frac{y_a + y_b}{2} - \frac{y_c - y_a}{x_c - x_a}\frac{y_a + y_c}{2}}{\frac{y_b - y_a}{x_b - x_a} -\frac{y_c - y_a}{x_c - x_a}} $$

Answer 4

Outline:

Let $ \vec{A}=\langle{a_1,a_2}\rangle, \vec{B}=\langle{b_1,b_2}\rangle, \vec{C}=\langle{c_1,c_2}\rangle$.

Find the midpoints (as vectors) for two of the sides (any two).

So for example, let $\vec{M}={\Large{\frac{A+B}{2}}}$ and let $\vec{N}={\Large{\frac{B+C}{2}}}$.

Let the circumcenter be represented as $\vec{P}=\langle{x,y}\rangle$, where $x,y$ are unknowns.

We have $2$ unkonwns $x,y$ but we have $2$ (linear) equations $$ \vec{PM}{\,\cdot\,}\vec{AB}=0\\ \vec{PN}{\,\cdot\,}\vec{BC}=0\\ $$ so you can solve for $x,y$.

Give it a try.