Another question of coordinate geometry of circles

Hi, I have another question: Find the equation of the circle passing through the points A(6, 3), B(-5, 2) and C(7, 2).

I understand the method of how to find it, in that 2 lines are made from these points and you have to find the perpendicular bisector of each and then make a simultaneous equation out of the perpendicular bisectors. I know that after you've solved that simultaneous equation and found the x coordinate for the centre of the circle, you substitute that in to find y, and then you use pythagoras' theorem to find the radius. I've attempted the question, but I got it wrong. The answer is

(x-1)² + (y+3)² = 61

This is some of my working out:

Midpoint of AB = (-0.5, 2.5)

Gradient of AB = 11, Gradient of perpendicular bisector to AB = -1/11

y-(2.5) = -1/11(x--0.5)

y = -1/11x + 27/11

Midpoint of BC = (-0.5, 0.5)

Gradient of BC = -0.5, Gradient of perpendicular bisector BC = 2

y-(0.5) = 2(x--0.5)

y = 2x -0.5

2x -0.5 = -1/11x + 27/11

I got x = 2/3, but x = -1, so I would really appreciate it if someone could point out the mistakes I've made.