Question #119454
Is there a sphere passing through (4,0,1),(10,-4,9),(-5,6,-11)and (1,2,3)?If so,find its equation.
1
Expert's answer
2020-06-02T19:15:16-0400

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

(xa)2+(yb)2+(zc)2=r2,a,b,c,rR(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:

{(4a)2+b2+(1c)2=r2(10a)2+(4+b)2+(9c)2=r2(5+a)2+(6b)2+(11+c)2=r2(1a)2+(2b)2+(3c)2=r2\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:

{12a8b+16c=18018a+12b20c=1656a4b4c=3\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=17424c=174 . This implies c=7.25c=7.25. Now we multiply the last equation by 3 and add to the second. We receive 32c=174-32c=174. Then c=5.4375c=-5.4375 . Thus, we received different values of cc. 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).


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Comments

Assignment Expert
07.06.20, 22:34

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Beenu. Verma
05.06.20, 16:31

Thank you so much

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