using the knowledge of systems of equations, and that the equation of a circle is (x-h)^2 + (y-k)^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.)

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

To find the equation of a circle that contains the points (6,2), (4,6), and (-3,5), we need to follow these steps:

Step 1: Write down the equation for each of the three given points.

Let's take the point (6,2). The equation of a circle can be written as (x-h)^2 + (y-k)^2 = r^2, where (h, k) is the center and r is the radius. Plugging in the coordinates of the point (6,2), we get:
(6-h)^2 + (2-k)^2 = r^2

Similarly, for the second point (4,6), the equation will be:
(4-h)^2 + (6-k)^2 = r^2

And for the third point (-3,5), the equation will be:
(-3-h)^2 + (5-k)^2 = r^2

Step 2: Rewrite the three equations to a standard form.

Expand each equation to get rid of the squared terms:

(6-h)(6-h) + (2-k)(2-k) = r^2
(4-h)(4-h) + (6-k)(6-k) = r^2
(-3-h)(-3-h) + (5-k)(5-k) = r^2

Simplifying further:

36 - 12h + h^2 + 4 - 4k + k^2 = r^2
16 - 8h + h^2 + 36 - 12k + k^2 = r^2
9 + 6h + h^2 + 25 - 10k + k^2 = r^2

Step 3: Rearrange the equations to form a system of equations.

Rearrange the equations by grouping the similar terms together:

h^2 - 12h - 4k + k^2 = r^2 - 40
h^2 - 8h - 12k + k^2 = r^2 - 52
h^2 + 6h - 10k + k^2 = r^2 - 34

Step 4: Solve the system of equations using elimination.

To eliminate one variable, we can subtract the first equation from the second equation and the first equation from the third equation.

(h^2 - 8h - 12k + k^2) - (h^2 - 12h - 4k + k^2) = (r^2 - 52) - (r^2 - 40)
(h^2 + 6h - 10k + k^2) - (h^2 - 12h - 4k + k^2) = (r^2 - 34) - (r^2 - 40)

Simplifying further:

4h + 8k = -12
18h - 6k = 6

To eliminate the variable 'k', we can multiply the first equation by 6 and the second equation by 8:

24h + 48k = -72
144h - 48k = 48

Adding these two new equations together, we can eliminate the variable 'k':

24h + 144h = -72 + 48

Simplifying further:

168h = -24

Dividing both sides by 168:

h = -24/168
h = -1/7

Substituting this value of 'h' into one of the original equations, we can solve for 'k':

4(-1/7) + 2k = -40

Simplifying further:

-4/7 + 2k = -40

Adding 4/7 to both sides:

2k = -40 + 4/7

Simplifying further:

2k = -280/7 + 4/7

Combining the fractions:

2k = -276/7

Dividing both sides by 2:

k = -276/14
k = -138/7

So, the center of the circle is (h, k) = (-1/7, -138/7), and the radius can be found by substituting this value into any of the original equations and solving for 'r'.