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 :)

Algebra ->  Length-and-distance -> 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 :)       Log On


   

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) About Me  (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,
x%5E2%2Bfx%2By%5E2%2Bgy%2Bh=0
Input each point,
4%2B2f%2B36%2B6g%2Bh=0
1.2f%2B6g%2Bh=-40
.
.
9-3f%2B1%2Bg%2Bh=0
2.-3f%2Bg%2Bh=-10
.
.
64%2B8f%2B36-6g%2Bh=0
3.8f-6g%2Bh=-100
Now subtract eq. 1 from eq. 2 to eliminate h,
-3f%2Bg%2Bh-2f-6g-h=-10%2B40
-5f-5g=30
4.f%2Bg=-6
.
.
.
8f-6g%2Bh-2f-6g-h=-100%2B40
6f-12g=-60
5.f-2g=-10
Subtract eq. 4 from eq. 5,
f-2g-f-g=-10-%28-6%29
-3g=-4
g=4%2F3
Then,
f%2B4%2F3=-6
f=-18%2F3-4%2F3
f=-22%2F3
and finally,
-3f%2Bg%2Bh=-10
22%2B4%2F3%2Bh=-10
h=-30%2F3-4%2F3-66%2F3
h=-100%2F3
.
.
.
x%5E2-%2822%2F3%29x%2By%5E2%2B%284%2F3%29y-100%2F3=0
So now complete the square to find the center (h,k).

%28x-11%2F3%29%5E2%2B%28y%2B2%2F3%29%5E2=300%2F9%2B121%2F9%2B4%2F9
%28x-11%2F3%29%5E2%2B%28y%2B2%2F3%29%5E2=425%2F9
The center is (11%2F3,-2%2F3).

Answer by Alan3354(69443) About Me  (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.