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

Points: (6, 3) (-5, 2) (7, 2)

Substitute the points into the general equation to form simultaneous equations:-

!2g+6f+c = -45

-10g+4f+c = -29

14g+4f+c = -53

Solving the above simultaneous equations: g = -1, f = 3 and c = -51

Therefore: x^2 -2x +y^2 +6y -51 = 0

Completing the squares of x and y: (x-1)^2 + (y+3)^2 -1-9-51 = 0

So it follows: (x-1)^2 + (y+3)^2 = 61

Centre of the circle is at (1, -3) and its radius is the square root of 61

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Q: What is the centre and its radius of a circle that passes through the points of 6 3 and -5 2 and 7 2 on the Cartesian plane showing all work from start to finish?
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