SOLUTION: I need to find the point that is equidistant from (2,6) (-3,1) (8,-6). And I request that you use only formulas instead of drawing graphs. Thank you :)

Question 950791: I need to find the point that is equidistant from (2,6) (-3,1) (8,-6). And I request that you use only formulas instead of drawing graphs. Thank you :)
Found 2 solutions by Fombitz, Alan3354:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Assume that this point (a,b) is the center of a circle and the points lie on the circle.
The circle has equation,

Input each point,

1.
.
.

2.
.
.

3.
Now subtract eq. 1 from eq. 2 to eliminate ,

4.
.
.
.

5.
Subtract eq. 4 from eq. 5,

Then,

and finally,

.
.
.

So now complete the square to find the center (h,k).

The center is (,).

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I need to find the point that is equidistant from (2,6) (-3,1) (8,-6).
=============
Another way to do it:
Label the points A(2,6) B(-3,1) C(8,-6)
Find the perpendicular bisectors of AB and of BC.
That will give you 2 equations in x & y.
Find the solution of those 2 equations.
That's the center of the circle, and is equidistant from A, B & C.
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