Dear Uncle Colin,

I know a 3D circle passes through $A(4,-4,5)$, $B(0,4,1)$ and $C(0,0,5)$ and I need to find its centre and radius. I could do it in 2D, but I’m a bit stuck here!

Hi, CIRCLE, and thanks for your message!

There are, as usual, several ways to approach this. My instinctive approach:

• Find the plane the three points lie on
• Find the plane that bisects two of the points
• Find the plane that bisects another pair of points
• Where the three points meet is the centre of the circle.

### The plane of the three points

Traditionally, you’d find the equation of the plane by finding the normal vector: $\vec{BA} = \colvecthree{4}{-8}{4}$ and $\vec{BC} = \colvecthree{0}{-4}{4}$; the cross product of those is $\colvecthree{16}{16}{16}$, which (dividing by 16) gives a normal vector of $\colvecthree{1}{1}{1}$. The equation of the plane is then $\colvecthree{1}{1}{1} \cdot \colvecthree{x}{y}{z} = 5$.

This works for all three points, so we’re happy.

### Bisecting planes

We can find the plane that bisects $A$ and $C$ quite simply - the vector between them is $\colvecthree{-4}{4}{0}$ and their midpoint is $\colvecthree{2}{-2}{5}$.

That makes their bisecting plane $\colvecthree{1}{-1}{0} \cdot \colvecthree{x}{y}{z} = 4$.

The bisector of $B$ and $C$ isn’t much harder; the vector between them is $\colvecthree{0}{-4}{4}$ and their midpoint is $\colvecthree{0}{2}{3}$, so their bisecting plane is $\colvecthree{0}{-1}{1} \cdot \colvecthree{x}{y}{z} = 1$.

### Where they intersect

We have three equations:

• $x + y + z = 5$
• $x - y = 4$
• $-y + z = 1$

We can rearrange the last one to eliminate $z$ as $1+y$:

• $x + 2y = 4$
• $x - y = 4$

So $x=4$ and $y=0$, by inspection, putting the centre at $(4,0,1)$.

How far is this point from $\colvecthree{0}{0}{5}$? It’s $\sqrt{ 4^2 + 0^2 + 4^2} = 4\sqrt{2}$.

### Another way

We can work out the lengths of the sides as $|AB| = 4\sqrt{6}$ and $|AC|=|BC| = 4\sqrt{2}$.

That means we have an isosceles triangle! The centre of the circle lies on the line connecting $C$ to the midpoint of $A$ and $B$ (which is at $\colvecthree{2}{0}{3}$.

That line is $\vec{r} = \colvecthree{0}{0}{5} + \lambda\colvecthree{2}{0}{-2}$.

We also know that the diameter of the triangle’s circumcircle is $\frac{a}{\sin(A)}$. We can easily work out any of the angles using the cosine rule - for example, $\cos(A) = \frac{32 + 96 - 32}{2\times (4\sqrt{6}) \times (4\sqrt{2})} = \frac{96}{64\sqrt{3}} = \sqrt{3}{2}$. That points to $A = \piby6$.

The diameter of the circles is then $\frac{4\sqrt{2}}{\sin\left(\piby6\right)}$, which makes the radius $4\sqrt{2}$.

And the point on the line $4\sqrt{2}$ from $C$ is when $\lambda = 2$ - it’s $\colvecthree{4}{0}{1}$.

Hope that helps!

- Uncle Colin