Posted by Kevin on .
using the knowledge of systems of equations, and that the equation of a circle is (xh)^2 + (yk)^2 = r^2, where (h,k) is the center and r is the radius. Find the equation of a circle that contains the following points: (6,2), (4,6), and (3,5). (These 3 points each create an equation. Then you have a system. Use ELIMINATION to solve.)

College Algebra 
Reiny,
The method suggested for your question is not the best one in this case.
I will use the property that the centre lies on the intersection of the right  bisector of any two chords.
for (6,2) and (4,6)
slope = (62)/46) = 2
so slope of perpendicular is +1/2
midpoint of chord = ( (6+4)/2 , (2+6)/2 ) = (5,4)
equation of rtbisector : y =(1/2)x + b
but (5,4) lies on it
4 = (1/2)(5) + b
b = 45/2 = 3/2
b = 3/2 > y = (1/2)x +3/2
for (6,2) and (3,5)
slope of line = (52)/(36) = 1/3
slope of perp is 3
midpoint of chord = (3/2, 7/2)
y = 3x + b
but (3/2,7/2) lies on it
7/2 = 3(3/2) + b
b = 7/2  9/2 = 1  y = 3x  1
intersection: 3x1 = (1/2)x+3/2
times 2
6x  2 = x + 3
5x = 5
x = 1
then y = 3(1)  1 = 2
so we know the centre is (1,2)
using (6,2),
r^2 = (61)^2 + (22)^2
= 25
circle equation: (x1)^2 + (y2)^2 = 25
check:
for (6,2)  LS = 25 + 0 = 25✔
for (4,6)  LS = 9 + 16 = 25✔
for (3,5)  LS = 16 + 9 = 25✔ 
College Algebra 
Steve,
However, just to see how elimination works, plug in the points and solve the 3 equations. The trick is how to get rid of those pesky squared terms. Just equate the various values of r^2.
(6h)^2 + (2k)^2 = r^2
(4h)^2 + (6k)^2 = r^2
(3h)^2 + (5k)^2 = r^2
(6h)^2 + (2k)^2 = (4h)^2 + (6k)^2
(6h)^2 + (2k)^2 = (3h)^2 + (5k)^2
(4h)^2 + (6k)^2 = (3h)^2 + (5k)^2
3612h+h^2 + 44k+k^2 = 168h+h^2 + 3612k+k^2
3612h+h^2 + 44k+k^2 = 9+6h+h^2 + 2510k+k^2
168h+h^2 + 3612k+k^2 = 9+6h+h^2 + 2510k+k^2
3612h + 44k = 168h + 3612k
3612h + 44k = 9+6h + 2510k
168h + 3612k = 9+6h + 2510k
4h +8k = 12
18h + 6k = 6
14h  2k = 18
Since we only have h and k, we only need two equations:
h + 2k = 3
6h + 2k = 2
5h = 5
h = 1
k = 2
(x1)^2 + (y2)^2 = r^2
pick any point:
(61)^2 + (22)^2 = 25 = r^2
r = 5
(x1)^2 + (y2)^2 = 25