Equation of Circles

(a) Let M be the mid-pt of PQ.
M = ((2+10)/2 , 0) = (6,0)
Hence radius = OM = 6
By perpendicular bisector of a chord, the centre C has same x-coordinate as M, then let C = (6,c)
Radius = √[(6-2)^2 + c^2] = √(c^2 + 16) = 6
√(c^2 + 16) = 6
c^2 + 16 = 36
c^2 = 20
c = -2√5 (c < 0)
Centre C = (6, -2√5)

(b) Equation of the circle:
(x - 6)^2 + (y + 2√5)^2 = 6^2
x^2 - 12x + 36 + y^2 + 4√5 y + 20 = 36
x^2 + y^2 - 12x + 4√5 y + 20 = 0