Finding center of circle from 3 coordinates How would I solve this question (from the SAT):

In the coordinate plane,the points $F (-2,1)$, $G (1,4)$, and $H (4,1)$ lie on a circle with center P. What are the coordinates of point P ?
$a)   (0,0)$
$b)   (1,1)$
$c)   (1,2)$
$d)   (1,-2)$
$e)   (2.5,2.5)$

I don't know how to find it quickly.
I can only find it through making $3$ equations in the form $(x-b)^2 + (y-h)^2 = r^2$.
Is there a fast way to find this? The question is only level. I am guessing they are assuming that point $F$ and $H$ are directly opposite in the circle but this is not stated explicitly.
 A: Since this is a Multiple Choice question and since all the coordinates are nice whole numbers, then I would say just plot all the answers and see which one works.
A: As Stephen Douglas Allen commented, $F$ and $H$ are horizontally aligned and the center should be on the perpendicular bissector; hence, its abscissa is $(-2+4)/2=1$ (that's probably why most suggested answers have 1 as first coordinates...).
And $H$ and $G$ are symmetric with respect to $y=x$, so the center should be on the first bisector. In other words, both coordinates should be equal. Note that this observation simply stems from the fact that $(1,4)$ is the reverse of $(4,1)$; no need for a drawing.
In the end, the center is at the intersection of both: $(1,1)$.
A: I would take a somewhat more "brute-force" (and arguably faster) method to this. Recall a set a points all share the same distance to a common point if and only if they all lay on a circle around that point. With that logic, it's simply a matter of checking distances.
(a)
$d((0,0), (-2,1))=\sqrt{(-2)^2+1^2}=\sqrt{5}$
$d((0,0), (1,4))=\sqrt{1^2+4^2}=\sqrt{17}\neq\sqrt{5}$
Not that one...
(b)
$d((1,1), (-2,1))=\sqrt{(1+2)^2+(1-1)^2}=\sqrt{9}=3$
$d((1,1), (1,4))=\sqrt{(1-1)^2+(1-4)^2}=\sqrt{9}=3$
$d((1,1), (4,1))=\sqrt{(1-4)^2+(1-1)^2}=\sqrt{9}=3$
Found it. The answer is (b). You can verify the rest of the answers don't work, if you have time.
