Find the center of the circle through the points $(-1,0,0),(0,2,0),(0,0,3).$
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.
Best Answer
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, $\displaystyle\;D = \frac{26 A + 80 B + 90C}{26 + 80 + 90} = \left(-\frac{35}{98},\frac{40}{49},\frac{135}{98} \right).$