In $\mathbb{R}^3$ find the equation of the circle passing for three points.

Question

So, I have the following points: $\left( \begin{matrix} 5 \\ 0 \\ 0 \end{matrix} \right), \left( \begin{matrix} 0 \\ 4 \\ -1 \end{matrix} \right), \left( \begin{matrix} -4 \\ 4 \\ 3 \end{matrix} \right)$ and I need to find the equation of the circle passing through them.

Here is how I solved it, but it was very long and tiring. Can you think of a smarter and easier way?

Thank you very much for your kind help!

Answer 1

Here is another approach:

Let the points be $x_1,x_2,x_3 \in \mathbb{R}^3$. Let $d = (x_1-x_3) \times (x_2-x_3)$, then $d$ is a normal to the plane containing the $x_i$, let $\alpha = \langle d, x_1 \rangle$, then the $x_i$ all lie on the plane $H = \{ x | \langle d, x \rangle = \alpha \}$.

Let $c$ be centre of the circle. It follows that $c \in H$, or in other words, $\langle d, c \rangle = \alpha$.

All the $x_i$ are equidistant from the centre, so we have $\|x_i-c\| = \|x_j -c\|$ for all $i,j$. These equations can be written as $\|x_1-c\|^2 = \|x_3 -c\|^2$ and $\|x_2-c\|^2 = \|x_3 -c\|^2$. Expanding the first of these equations gives $\|x_1\|^2-2 \langle x_1, c \rangle + \|c\|^2 = \|x_3\|^2-2 \langle x_3, c \rangle + \|c\|^2$, which simplifies to $\langle x_1-x_3,c \rangle = \frac{1}{2}(\|x_1\|^2-\|x_3\|^2)$. Similarly for the second equation.

Gathering these equations gives \begin{eqnarray} d^T c &=& \alpha \\ (x_1-x_3)^T c &=& \frac{1}{2}(\|x_1\|^2-\|x_3\|^2) \\ (x_2-x_3)^T c &=& \frac{1}{2}(\|x_2\|^2-\|x_3\|^2) \\ \end{eqnarray} As long as the $x_i$ are not collinear, this set of linear equations have a unique solution which is the centre of the circle. The sphere containing the $x_i$ whose centre lies on $H$ is given by $\|x-c\|^2 = \|x_1-c\|^2$.

Actually doing the computations gives: $$ \begin{bmatrix} 16 & 24 & 16 \\ 9 & -4 & -3 \\ 4 & 0 & -4\end{bmatrix} c = \begin{bmatrix} 80 \\ -8 \\ -12 \end{bmatrix} $$ which gives $c=\frac{1}{34} (19, 20, 121)^T$, and the radius squared is $r^2=\|x_1-c\|^2 = \frac{1113}{34}$.

Answer 2

I noticed the points are linearly indipendent so I can use them as a base. With such a base the equation I need is $X_1^2+X_2^2+X_3^2=1$. In matrix form the equations is $X^t I_3 X = 1$ where $X=(X_1,X_2,X_3)$.

We now indicate with $Y=(Y_1,Y_2,Y_3)$ the coordinates in the canonical base. Let's call $P=\left( \begin{matrix} 5 & 0 & -4 \\ 0 & 4 & 4 \\ 0 & -1 & 3 \end{matrix} \right)$. Then we have $Y=XP \Rightarrow X = Y P^{-1}$.

Then we find the equation by solving $Y^t(P^{-1})^t I_3 (P^{-1})Y=1$.

Now, I also solved it and found the equation but it was quite painful, also because this is the result: $Y_1^2+\frac{1}{2} Y_1 Y_2+\frac{133}{128} Y_2^2+2 Y_1 Y_3 - \frac{17}{16}Y_2 Y_3 + \frac{33}{8} Y_3^2=25$

Answer 3

Another way to solve the problem is by determining the center of the circle. First, determine the plane containing the three given points, hopefully you will find $$2X+3Y+2Z=10.$$ Next, find parametrisations for the perpendicular bisectors of two pairs of points on the circle. The perpendicular bisector of $(0,4,-1)$ and $(-4,4,3)$ is the line pasing through the point $$\frac{1}{2}\cdot\big((0,4,-1)+(-4,4,3) \big)=(-2,4,1),$$ and perpendicular to the line through $(0,4,-1)$ and $(-4,4,3)$. Then it is perpendicular to $(0,4,-1)-(-4,4,3)=(4,0,-4)$, and one such vector in our plane is $(3,-4,3)$. So a parametrisation of the perpendicular bisector of $(0,4,-1)$ and $(-4,4,3)$ is $$(-2,4,1)+a\cdot(3,-4,3).$$ Similarly, a parametrisation of the perpendicular bisector of $(0,4,-1)$ and $(5,0,0)$ is $$\frac{1}{2}(5,4,-1)+b\cdot(11,8,-23).$$ These two lines intersect at $C=\left(\frac{19}{34},\tfrac{10}{17},\tfrac{121}{34}\right)$ for $a=\tfrac{29}{34}$ and $b=-\tfrac{3}{17}$, which is thus the center of the circle. Its radius is then the distance from $C$ to $(5,0,0)$, which is $$\sqrt{\left(\frac{19}{34}-5\right)^2+\left(\frac{10}{17}\right)^2+\left(\frac{121}{34}\right)^2}=\sqrt{\frac{1113}{34}},$$ so an equation for the circle passing through the three given points is $$\left(X-\frac{19}{34}\right)^2+\left(Y-\frac{10}{17}\right)^2+\left(Z-\frac{121}{34}\right)^2=\frac{1113}{34},$$ or after clearling out denominators, $$(34X-19)^2+(34Y-20)^2+(34Z-121)^2=34\cdot1113.$$