# 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

1. $5,6,1$
2. $5,6,-1$
3. $5,-6,1$
4. $-5,-6,-1$

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

• You could alternatively use that the angle in a semicircle is a right angle. May 6 '12 at 16:10
• thanks for the help but i would like to solve it using 3-d
– Ashu
May 6 '12 at 16:18
• Mark's solution applies both in two and three dimensions. May 6 '12 at 16:23
• You need three points to define a circle, so you can't do that from the two you are given - so you need to identify a sphere. You can test for vectors being perpendicular by taking a dot product, which is easy arithmetic. May 6 '12 at 16:25
• It looks like none of the points 1-4 are on the sphere. May 6 '12 at 16:28

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 $P = (5,6,-1)$.

• thanks great answer... seems quiet fast..
– Ashu
May 7 '12 at 5:39
• @Ashu: Glad to help. May 7 '12 at 6:05

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.

• this solution is right for my case but what if the center was $(0,0,0)$
– Ashu
May 6 '12 at 16:21
• If the centre turned out to be $(0,0,0)$, you should feel lucky, it makes computations easier. Unfortunately, that is not the case here. The centre of the circle is always the midpoint of any line segment that joins the endpoints of (any) diameter. May 6 '12 at 16:24
• i mean that if the center was 0,0,0 then all the dist of the 4 pts would be the same as during squaring the signs are removed
– Ashu
May 6 '12 at 16:42
• True. But no one would ask a multiple choice question in which all choices were correct. Though come to think of it, that might be a way to avoid complaints about low grades. May 6 '12 at 16:45
• yeah thats true ... any ways i think your method would be a little too long (the time limit for solving is just one minute). what if the correct ans is D...checking all the point for the radius will take a little long
– Ashu
May 6 '12 at 16:55

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