What is the standard equation of the circle that passes through the points (-2,4),(3,-1),(6,8)?

The equation of a circle is (x - a)^2 + (y - b)^2 =
r^2

As we know three points through which the circle
passes, we can create three equations.

(-2 ,
4)

(-2 - a)^2 + (4 - b)^2 = r^2
...(1)

(3 , -1)

(3 - a)^2 +
(-1 - b)^2 = r^2 ...(2)

(6 ,
8)

(6 - a)^2 + (8 - b)^2 = r^2
...(3)

(1) - (2)

=> (-2
- a)^2 + (4 - b)^2 - (3 - a)^2 - (-1 - b)^2 = 0

=>
(-2 - a - 3 + a)(-2 - a + 3 - a) + (4 - b - 1 - b)(4 - b + 1 +b) =
0

=> -5(1 - 2a) + (3 - 2b)(5) =
0

=> 2a - 1 + 3 - 2b =
0

=> a - b + 1 = 0

(2)
- (3)

=> (3 - a)^2 + (-1 - b)^2 - (6 - a)^2 - (8 -
b)^2 = 0

=> (3 - a + 6 - a)(3 - a - 6+ a) + ( - 1 -
b - 8 + b)(-1 - b + 8 - b) =0

=> (9 - 2a)(-3) + -9(
7 - 2b) = 0

=> 9 - 2a + 21 - 6b =
0

=> 2a + 6b - 30 =
0

=> a + 3b - 15 =
0

Using a - b + 1 =
0

=> a = b - 1

b - 1 +
3b - 15 = 0

=> -4b - 16 =
0

=> b = 4

a =
3

(-2 - a)^2 + (4 - b)^2 =
r^2

=> (-2 - 3)^2 + ( 4 - 4)^2 =
r^2

=> 5^2 =
r^2

=> r =
5

The equation of the circle is (x - 3)^2 +
(y - 4)^2 = 5^2