Answer to Question #119454 in Analytic Geometry for Beenu Varma

Question

Is there a sphere passing through (4,0,1),(10,-4,9),(-5,6,-11)and (1,2,3)?If so,find its equation.

Answer 1

We remind that an equation of a sphere has a form:

$(x-a)^2+(y-b)^2+(z-c)^2=r^2,\quad a,b,c,r\in{\mathbb{R}}$

After substitution of points we get a system of four equations:

$\left\{ \begin{matrix} (4-a)^2+b^2+(1-c)^2=r^2 \\ (10-a)^2+(4+b)^2+(9-c)^2=r^2 \\ (5+a)^2+(6-b)^2+(11+c)^2=r^2 \\ (1-a)^2+(2-b)^2+(3-c)^2=r^2 \\ \end{matrix}\right.$

We subtract the first equation from the others, simplify expressions and receive:

$\left\{ \begin{matrix} 12a-8b+16c=180 \\ -18a+12b-20c=165 \\ 6a-4b-4c=3 \\ \end{matrix}\right.$

We multiply the last equation by (-2) and add to the first one. Then we get $24c=174$ . This implies $c=7.25$. Now we multiply the last equation by 3 and add to the second. We receive $-32c=174$. Then $c=-5.4375$ . Thus, we received different values of $c$. Therefore,

the system has no solutions. Thus, there is no sphere passing through points (4,0,1),(10,-4,9),(-5,6,-11)and (1,2,3).