Find the center of the circle through the points $(-1,0,0),(0,2,0),(0,0,3).$

Question

Find the center of the circle through the points $(-1,0,0),(0,2,0),(0,0,3).$

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Let the circle passes through the sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ and the plane $Ax+By+Cz+D=0$

So the equation of the circle is $x^2+y^2+z^2+2ux+2vy+2wz+d+\lambda(Ax+By+Cz+D)=0$

As it passes through $(-1,0,0)$ so

$1-2u+d+\lambda(-A+D)=0$

As it passes through $(0,2,0)$ so

$4+4v+d+\lambda(2B+D)=0$

As it passes through $(0,0,3)$ so

$9+6w+d+\lambda(3C+D)=0$

Here i am stuck.I cannot find the radius of the circle.Is my method correct or not?Is there simpler method possible?Please help.

Answer 1

Let $\Omega(x_0,y_0,z_0)$ be the center of this circle, and let us denote our points by $A,B$ and $C$, respectively. Then, by definition, we have the following family of equalities: $$\begin{cases}\operatorname{dist}(\Omega,A)=\operatorname{dist}(\Omega,B)\\ \operatorname{dist}(\Omega,C)=\operatorname{dist}(\Omega,B)\\ \operatorname{dist}(\Omega,C)=\operatorname{dist}(\Omega,A) \end{cases}\iff\begin{cases}(x_0+1)^2+y_0^2+z_0^2=x_0^2+(y_0-2)^2+z_0^2\\ x_0^2+y_0^2+(z_0-3)^2=x_0^2+(y_0-2)^2+z_0^2\\ x_0^2+y_0^2+(z_0-3)^2=(x_0+1)^2+y_0^2+z_0^2 \end{cases}$$ If we add the condition that $\Omega$ is in the plane formed by these three points, then this should be sufficient to determine the coordinates of the center of the circle.

Answer 2

follow this process

1. find the plane passing through the three points (as the 3 points are points on axis it is very simple)
2. take normal vector to the plane
3. consider any two sides of the triangle made by the three points and take their mid-point
4. now take cross product between side's vector and plane's normal vector
5. this will give you direction of perpendicular to the side
6. now take a line passing through the mid-point of that side
7. do the same process for the second side
8. the point these both points intersect is your center of the circle

we just intersected two perpendicular bisectors of the sides

Answer 3

To express that the center is equidistant from two points, you write

$$(x_c-x_0)^2+(y_c-y_0)^2+(z_c-z_0)^2=(x_c-x_1)^2+(y_c-y_1)^2+(z_c-z_1)^2.$$

This can be rewritten as

$$(2x_c-x_0-x_1)(x_1-x_0)+(2y_c-y_0-y_1)(y_1-y_0)+(2z_c-z_0-y_1)(z_1-z_0)=0,$$ which is linear in $x_c,y_c,z_c$. This is in fact the equation of the bissector plane of the two points.

By considering the equations of two bissectors and that of the plane through the given points, you get a $3\times3$ linear system.

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For efficient computation, you can translate all points so that $P_0$ comes to the origin. Then

$$\left(\begin{matrix}x'_1&y'_1&z'_1\\x'_2&y'_2&z'_2\\y'_1z'_2-y'_2z'_1&z'_1x'_2-z'_2x'_1&x'_1y'_2-x'_2y'_1\end{matrix}\right) \left(\begin{matrix}2x'_c\\2y'_c\\2z'_c\end{matrix}\right)= \left(\begin{matrix}{x_1'}^2+{y_1'}^2+{z_1'}^2\\{x_2'}^2+{y_2'}^2+{z_2'}^2\\0\end{matrix}\right).$$

Answer 4

You essentially want to find the circumcenter $D$ of the triangle $\triangle_{ABC}$ with vertices $$A = (-1,0,0), B = (0,2,0), C = (0,0,3)$$

Since $D$ is lying on the plane holding $A, B, C$, there exists $3$ real numbers $\alpha,\beta,\gamma$ such that

$$D = \alpha A + \beta B + \gamma C\quad\text{ with }\quad \alpha + \beta + \gamma = 1$$ This $3$-tuple is called the baricenteric coordinate of $D$. They can be computed using the sides $a,b,c$ of the triangle alone:

$$\alpha : \beta : \gamma \;=\; a^2(-a^2 + b^2 + c^2) : b^2( a^2 - b^2 + c^2) : c^2(a^2 + b^2 - c^2)$$

For the triangle at hand, we have $(a^2,b^2,c^2) = (13,10,5)$. This leads to $$\alpha : \beta : \gamma \;=\; 13(-13+10+5) : 10(13-10+5) : 5(13+10-5) = 26 : 80 : 90$$ As a result, $\displaystyle\;D = \frac{26 A + 80 B + 90C}{26 + 80 + 90} = \left(-\frac{35}{98},\frac{40}{49},\frac{135}{98} \right).$