Question 265032
the solution is to find the equations of the
perpendicular bisectors of any 2 of the chords
formed by these 3 points.
The chords are:
AB
AC
BC
For the chord A(-4,0) to C(6,0):
{{{P[x] = (-4 + 6)/2}}}
{{{P[x] = 1}}}
{{{P[y] = (0 + 0)/2}}}
{{{P[y] = 0}}}
The center of this chord is at P(1,0)
For the chord from B(4,8) to C(6,0):
{{{P[x] = (4 + 6)/2}}}
{{{P[x] = 5}}}
{{{P[y] = (8 + 0)/2}}}
{{{P[y] = 4}}}
The center of this chord is at P(5,4)
Now I have to find the slopes of these 2 chords
that I picked. 
For the chord AC:
{{{m = 0/(6 - (-4))}}}
{{{m = 0}}}
For the chord BC:
{{{m = (8 - 0)/(4 - 6)}}}
{{{m = 8/(-2)}}}
{{{m = -4}}}
Now I need to find perpendicular bisectors of the chords
Any line perpendicular to AC will be parallel to the y-axis
since {{{m=0}}} is parallel to the x-axis.
And I also know the line goes through P(1,0).
This is just the line 
(1) {{{x = 1}}}
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Any line perpendicular to BC will have slope = {{{-(1/m)}}}
{{{-(1/(-4)) = 1/4}}}
And this line goes through P(5,4)
The formula to use is:
{{{(y - y[1])/(x - x[1]) = m}}}
{{{(y - 4)/(x - 5) = 1/4}}}
{{{y - 4 = (x - 5)/4}}}
{{{4y - 16 = x - 5}}}
{{{4y = x + 11}}}
(2) {{{y = (1/4)*x + (11/4)}}}
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Now I just need to know where lines
(1) and (2) intersect
(1) {{{x = 1}}}
(2) {{{y = (1/4)*x + (11/4)}}}
{{{y = (1/4)*1 + 11/4}}}
{{{y = 12/4}}}
{{{y = 3}}}
The 2 lines intersect at
the center of the circle at (1,3)
I'll try to plot line {{{y = (1/4)*x + 11/4}}} to check
{{{ graph(500,500,-10,10,-10,10,(1/4)*x + 11/4 )}}}
Looks like the line intersects {{{x = 1}}} at (1,3)