Circle o, with center (x, y), passes through the points a(0, 0), b(–3, 0), and c(1, 2). find the coordinates of the center of the circle. in your final answer, include all formulas and calculations used to find point o, x, y), the center of circle o.

The

the main formula of the circle's equation is 
(x-a)²+ (y-b)² = R²
where C(a, b) is the center of the circle
R is the radius

if a point A(x', y') passes through the circle, so the equation of the circle can be written as 
(x'-a)²+ (y'-b)² = R², and that is  a main formula.

Circle O, with center (x, y), passes through the points A(0, 0), B(–3, 0), and C(1, 2), so we have exactly three equation:

(0-x)² + (0-y)² = R², circle O passes through A
x²+y²= R²
(-3 -x)² + (0-y)² = R², circle O passes through B
(-3 -x)² + (y)² = R²
(1-x)² + (2-y)² = R², circle O passes through A
(1-x)² + (2-y)² = R²

and we know that R= OA = OC= OB, 
OA=R= sqrt( (0-x)² + (0-y)² ) = OB = sqrt((-3 -x)² + (0-y)²), this implies

x²+y² = (-3 -x)² + (0-y)² , it implies x² = 9+ x² + 6x , and then -9/6=x, x= -3/2
and when OA = OC
x²+y² =(1-x)² + (2-y)²  so, x²+y² =1+x²-2x +4+y²-4y, therefore -5= -2x -4y
 -5= -2x -4y, when x = -3 /2   we obtain y = 2

the center is C(-3/2, 2)

Center of circle = -3/2, 2

Step-by-step explanation:

Substitute the points into the equation

0^2 + 0^2 + 2a(0) + 2b(0) + c = 0

2a + 2b + c = 0

c = 0

Next point is (-3,0)

-3^2 + 2a(-3) + 2b + c = 0

9 - 6a + c = 0

-6a + c = -9

-9/-6 = a

a = 3/2

Now, put a and c into the format to find b

1^2 + 2^2 + 2(3/2)(1) + 2b(2) + 0 = 0

1 + 4 + 3 + 4b = 0

8 + 4b = 0

2 + b = 0

b = -2

-3/2, 2 = center of circle

Sin(a)+cos(a)+cot(a)-1=2(csc(x)+1)