3-D equation of a circle I came across a sum but could not solve it as i dont know the 3d equations of a circle :
The sum is   If $A(3,-2,2)$ and $B(2,9,5)$ are the end points of a diameter of a circle,then the third pt that lies on the circle is  


*

*$5,6,1$

*$5,6,-1$

*$5,-6,1$

*$-5,-6,-1$    


Ans:  

 2

Would appreciate if someone can explain the equation of a circle (whether parametric or a normal equation) before helping solve the sum.  
P.S- The time limit for solving the sum was one minute  
 A: To solve this problem, we can use quite basic tools.
$1$. Locate the centre $C$ of the circle. This is the midpoint of the line segment you were given.
$2$. Compute the radius $r$ of the circle.
$3$. Which one of the candidate points is at distance $r$ from $C$?
Once you know the coordinates $(c_1,c_2,c_3)$ of the centre $C$, and the radius $r$, then the equation of the sphere with centre $C$, radius $r$ is given by 
$$(x-c_1)^2+(y-c_2)^2+(z-c_3)^2=r^2.$$
A candidate point $P$ lies on a circle with the given line segment as a diameter if and only if it lies on the sphere with the above equation.
Or else we can use the Pythagorean Theorem. Find the square of the length of the diameter. For which of your candidate points $P$ is the sum of the squares of the distances from $P$ to the diameter ends equal to the square of the diameter? 
Or else we can use perpendicularity directly. Let our given points be $A$ and $B$, and let $P$ be a candidate point. For $P$ to lie on a circle that has $AB$ as a diameter, we need the "dot product" of $A-P$ and $B-P$ to be $0$. That might be fast enough for the $1$ minute restriction.
A: Let's run with @AndréNicolas' last suggestion, to use the dot product.
This seems to be the fastest way to solve this problem. 
Let $P = (5k_1,6k_2,k_3)$, where $k_i$ is $\pm 1$. 
This allows us to check all of the solutions at once. 
Notice that $k_i^2 = 1$. 
Then 
$$\begin{eqnarray*}
(A-P)\cdot(B-P) &=& A\cdot B - (A+B)\cdot P + P^2 \\
&=& (3,-2,2)\cdot(2,9,5) - (5,7,7)\cdot(5k_1,6k_2,k_3) + (5k_1,6k_2,k_3)^2 \\
&=& 60-25k_1-42k_2-7k_3. \\
&=& 0
\end{eqnarray*}$$
By inspection we see $(k_1,k_2,k_3) = (1,1,-1)$, so 
$P = (5,6,-1)$. 
A: hi this is hunde mine goes like this
let say
first it is in a x and y dimension 
x^2+y^2=r^2 
then rotate it in to z dimension along x axis
cos@=a/r)^2+y^2-z^2=r^2
this is hunde from Ethiopia 
