Circumcircle in 3D

Question

The vertices of a triangle are given as $A(1, 0, 0), B(3, 1, 0), C(5, 2, 1) $. Find the center and the radius of the circumcircle of this triangle.

My Attempt:

The normal to the plane of the triangle is

$ N = (B - A) \times (C - A) \\ = ( (3, 1, 0) - (1, 0, 0) ) \times ( (5, 2, 1) - (1, 0, 0) ) = (2, 1, 0) \times (4, 2, 1) = (1, - 2 , 0 ) $

The midpoint of $AB = (2, 0.5, 0)$ and the midpoint of $AC$ = (3, 1 , 0.5) $

The perpendicular bisector of $AB$ is given by

$ V_1 = N \times AB = (1, -2, 0) \times (2, 1, 0) = (0, 0, 5) $

And the perpendicular bisector of $AC$ is

$ V_2 = N \times BC = (1, -2, 0) \times (4, 2, 1) = (-2, -1, 10) $

Now we find the intersection of the two bisectors, for which we have to solve for $t$ and $s$ the following parametric vector equation

$ (2, 0.5, 0) + t (0, 0, 5) = (3, 1, 0.5) + s (-2, -1, 10) $

Dot multiplying the above equation by a vector perpendicular to $(0, 0, 5)$ such as $(0, 1, 0)$ eliminate $t$, and results in

$ 0.5 = 1 - s $

So $ s = 0.5 $

And dot multiplying by a vector perpendicular to $(-2, -1, 10)$ such as $(0, 10, 1)$ eliminate $s$, resulting in,

$ 5 + 5 t = 10.5 $

Hence $t = 1.1 $

Hence, the circumcenter is $(2, 0.5, 5.5) $

And from this, one can compute the circumradius which turns out to be $\sqrt{\dfrac{63}{2}} $

That was my method. I appreciate comments or alternative methods.

Answer 1

Another way to obtain the coordinates of the circumcenter is to get the intersection point of the plane $ABC$, the perpendicular bisector of $AB$ and the perpendicular bisector of $AC$.

The equation of the plane $ABC$ is

$\begin{vmatrix}x-x_A&y-y_A&z-z_A\\x_B-x_A&y_B-y_A&z_B-z_A\\x_C-x_A&y_C-y_A&z_C-z_A\end{vmatrix}=0\quad,$

$\begin{vmatrix}x-1&y&z\\2&1&0\\4&2&1\end{vmatrix}=0\quad,$

$x-2y-1=0\;.$

The equation of the perpendicular bisector of $AB$ is

$\left(x\!-\!x_A\!\right)^2\!+\!\left(y\!-\!y_A\!\right)^2\!+\!\left(z\!-\!z_A\!\right)^2\!=\!\left(x\!-\!x_B\!\right)^2\!+\!\left(y\!-\!y_B\!\right)^2\!+\!\left(z\!-\!z_B\!\right)^2\;,$

$\left(x-1\right)^2+y^2+z^2=\left(x-3\right)^2+\left(y-1\right)^2+z^2\;,$

$4x+2y-9=0\;.$

The equation of the perpendicular bisector of $AC$ is

$\left(x\!-\!x_A\!\right)^2\!+\!\left(y\!-\!y_A\!\right)^2\!+\!\left(z\!-\!z_A\!\right)^2\!=\!\left(x\!-\!x_C\!\right)^2\!+\!\left(y\!-\!y_C\!\right)^2\!+\!\left(z\!-\!z_C\!\right)^2\;,$

$\left(x-1\right)^2+y^2+z^2=\left(x-5\right)^2+\left(y-2\right)^2+\left(z-1\right)^2\;,$

$8x+4y+2z-29=0\;.$

Hence, we get the coordinates of the circumcenter by solving the following system of equations :

$\begin{cases}x-2y-1=0\\4x+2y-9=0\\8x+4y+2z-29=0\end{cases}$

By adding both sides of the first and the second equation, we obtain that

$5x-10=0\;\;$ that is $\;\;x=2\,.$

Consequently, from the first equation we obtain that $\;y=\dfrac12\,.$

Finally, from the third equation we obtain that $\;z=\dfrac{11}2\,.$

Therefore the coordinates of the circumcenter are :

$\left(2,\dfrac12,\dfrac{11}2\right)\,.$

Answer 2

Motivated by mine and Angelo's comments on Angelo's answer, another way you could find the solution to this would be to find the intersetion between a sphere and plane, both containing the three non-collinear points, $A(1, 0, 0), B(3, 1, 0), C(5, 2, 1) $.

As previously stated the equation for the plane defined by $A, B,C$ is (I'll denote it as $\alpha$) $$\alpha :x-2y-1=0$$

We know that the equation of a sphere is given as:
$(x-a)^2+(y-b)^2+(z-c)^2=R^2$, where $S(a,b,c)$ is its center and $R$ its radius. Now, by plugging the points $A,B,C$ in this equation for each point, respectively, we'll get the following system: $$(1-a)^2+(0-b)^2+(0-c)^2=R^2$$ $$(3-a)^2+(1-b)^2+(0-c)^2=R^2$$ $$(5-a)^2+(2-b)^2+(1-c)^2=R^2$$

This system has $4$ unknowns and $3$ equations, so that means there is an infinitely many spheres containing $A,B,C$.

Solving the system we can find that $a=a$, $b=\frac{1}{2}(9-4a)$, $c=\frac{11}{2}$ and $R=\frac{\sqrt{10a^2-40a+103}}{\sqrt{2}}$, $a \in \mathbb{R}$.
Now we can take $a=1$ ($a$ can be any real number and it will still give us the solution). Then the center of the sphere will be $S(1,2.5,5.5)$ and $R=\sqrt{\frac{73}{2}}$ its radius. So the sphere's equation is $$\mathscr{S}:(x-1)^2+(y-2.5)^2+(z-5.5)^2=\frac{73}{2}$$

The center of the circle in the intersection of sphere $\mathscr{S}$ and plane $\alpha$ will be the point of intersection between $\alpha$ and a line perpendicular to $\alpha$ that passes through center of the given sphere, $S$. It is simple shown that the equation of that line is:$$l:\frac{x-1}{1}=\frac{y-2.5}{-2}=\frac{z-5.5}{0}$$ Or given in the parametric form: $l:\begin{cases} x=t+1\\ y=-2t+2.5\\ z=5.5 \end{cases}, t \in \mathbb{R}$.

If we put these values of $x,y,z$ in $\alpha$, we get: $$t+1+4t-5-1=0\implies t=1$$

This means that the center of the circle you're looking for is $(2,0.5,5.5)$ (just like you've gotten).

Another way to find the center (which is unnecessarily complicated, compared to what you've probably done, finding the distance between center of circle and one of its points) is to find the distance between center of circle and center of sphere, $S$, then since we have radius of sphere, $R$, using Pythagorean theorem (where $R$ is hypotenuse) we can find the radius of the circle. I'll leave that as an exercise.