Question 2770
The equation of the circle is:
(x-h)^2 + (y-k)^2 = r^2
(h,k) are the coorcinates of the center of the circle
r is the radius
substituting the values given in the question in the equation of the circle we will ger:
(x-h)^2 + (y-6)^2 = 10^2
=> (x-h)^2 + (y-6)^2 = 100

Now if a circle passes through the point (x,y),
then
(x-h)^2 + (y-k)^2 - r^2 = 0
Therefore
(x-h)^2 + (y-6)^2 -100 =0....for (8,-2)
=> (8-h)^2 + (-2-6)^2 -100 =0
=> 64 -16h + h^2 + 64 - 100 = 0
=> h^2 -16h + 28 = 0
=> h^2 -2h -14h + 28 = 0
=> h(h-2) -14(h-2) = 0
=> (h-2)(h-14) = 0
=> According to the zero product rule
h-2 =0 or h-14 =0
=> h = 2 or 14