The general form of the equation of a circle is
(x-a)^2 + (y-b)^2 = r^2 where the center is (a,b)
Plugging in the given values:-
(6 - a)^2 + (4-b)^2 = r^2 and
(6-a)^2 + (2-b)^2 = r^2
Subtracting:-
(4-b)^2 - (2 - b)^2 = 0
16 - 8b + b^ - (4 -4b + b^2) = 0
12 - 4b = 0
-4b = -12
b = 3
Similarly if we take the 2 points (6,2) and G(10,2) and make two equations we can find the value of a
(6-a)^2 + (2-b)^2 = r^2
(10.a)^2 + (2-b)^2 = r^2 subtract::-
(6-a)^2 -(10-a(^2 = 0
this will give a = 8
so the center of the circle is the poit (8,3)
