The general equation of a circle is (x-h)^2+(y-k)^2=a^2, where (h,k) is the centre and a the radius.
If we put in the two points we get two equations:
(12-h)^2+(1-k)^2=a^2=(2-h)^2+(3+k)^2 [(-3-k)^2 is the same as (3+k)^2)]
Using the difference of two squares:
10(14-2h)+4(-2-2k)=0, 140-20h-8-8k=0, 132-20h-8k=0, 33-5h-2k=0, so k=(33-5h)/2.
(h,k) lies on the line 2x-5y+10=0, so 2h-5(33-5h)/2+10=0, 4h-5(33-5h)+20=0, 29h-145=0, so h=5.
Since h=5, k=(33-25)/2=4 and a^2=7^2+3^2=58.
The equation of the circle is: (x-5)^2+(y-4)^2=58.
Check: put (12,1) and (2,-3) into the equation. They check out because 9+49=49+9=58.
Put (5,4) in 2x-5y+10 and we get zero, which checks out.