A circle with centre (h,k) and radius a has the equation (x-h)^2+(y-k)^2=a^2. If we substitute x and y for the three points, we have three equations:
a) (-5,4)
(5+h)^2+(4-k)^2=a^2 [(-5-h)^2=(5+h)^2 because -1^2=+1^2=1]
b) (-3,8)
(3+h)^2+(8-k)^2=a^2=(5+h)^2+(4-k)^2
(3+h)^2-(5+h)^2+(8-k)^2-(4-k)^2=0
We have difference of squares twice on the left, so we can make use of A^2-B^2=(A-B)(A+B):
-2(8+2h)+4(12-2k)=0, -16-4h+48-8k=0, 32=4h+8k, 8=h+2k, so k=(8-h)/2.
c) (1,6)
(1-h)^2+(6-k)^2=a^2=(5+h)^2+(4-k)^2=a^2
6(-4-2h)+2(10-2k)=0, -24-12h+20-4k=0, -4-12h-4k=0, -1-3h-k=0=-1-3h-(8-h)/2, so 2+6h+8-h=0 and h=-2. So k=5.
Therefore, a^2=3^2+1=10.
The equation of the circle becomes (x+2)^2+(y-5)^2=10.