What is the centre and radius of a circle passing through the points of 2 2 and 1 1 and 7 -3 on the Cartesian plane showing work?

Answer 1

General equation of a circle on the Cartesian plane: x^2+y^2+2gx+2fy+c = 0

Points: (2, 2) (1, 1) and (7, -3)

Substitute the above values into the general equation to form simultaneous equations

4g+4f+c = -8

2g+2f+c = -2

14g-6f+c = -58

Solving the simultaneous equations: g = -4, f = 1 and c = 4

Substituting the above values into the general equation: x^2+y^2-8x+2y+4 = 0

Completing the squares: (x-4)^2+(y+1)^2-16-1+4 = 0

So: (x-4)^2+(y+1)^2 = 13

Therefore centre of circle is at (4, -1) and its radius is the square root of 13