Posted by Kevin on Thursday, December 15, 2011 at 12:27am.
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 = (6-2)/4-6) = -2
so slope of perpendicular is +1/2
midpoint of chord = ( (6+4)/2 , (2+6)/2 ) = (5,4)
equation of rt-bisector : y =(1/2)x + b
but (5,4) lies on it
4 = (1/2)(5) + b
b = 4-5/2 = 3/2
b = 3/2 --------> y = (1/2)x +3/2
for (6,2) and (-3,5)
slope of line = (5-2)/(-3-6) = -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: 3x-1 = (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 = (6-1)^2 + (2-2)^2
= 25
circle equation: (x-1)^2 + (y-2)^2 = 25
check:
for (6,2) --- LS = 25 + 0 = 25✔
for (4,6) --- LS = 9 + 16 = 25✔
for (-3,5) -- LS = 16 + 9 = 25✔
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.
(6-h)^2 + (2-k)^2 = r^2
(4-h)^2 + (6-k)^2 = r^2
(-3-h)^2 + (5-k)^2 = r^2
(6-h)^2 + (2-k)^2 = (4-h)^2 + (6-k)^2
(6-h)^2 + (2-k)^2 = (-3-h)^2 + (5-k)^2
(4-h)^2 + (6-k)^2 = (-3-h)^2 + (5-k)^2
36-12h+h^2 + 4-4k+k^2 = 16-8h+h^2 + 36-12k+k^2
36-12h+h^2 + 4-4k+k^2 = 9+6h+h^2 + 25-10k+k^2
16-8h+h^2 + 36-12k+k^2 = 9+6h+h^2 + 25-10k+k^2
36-12h + 4-4k = 16-8h + 36-12k
36-12h + 4-4k = 9+6h + 25-10k
16-8h + 36-12k = 9+6h + 25-10k
-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
(x-1)^2 + (y-2)^2 = r^2
pick any point:
(6-1)^2 + (2-2)^2 = 25 = r^2
r = 5
(x-1)^2 + (y-2)^2 = 25