What is the equation of the circle having a radius of square root of 85, through (5,9) and (1,-7)? In solving this problem, I first solved the slope of the perpendicular bisector and I solved the midpoint of the two points. I don't know what to do next. I hope you can help me.

Let center to be $(a,b)$. $$(a-5)^2+(b-9)^2=85, (a-1)^2+(b+7)^2=85$$
$$a^2-10a+b^2-18b=-21,a^2-2a+b^2+14b=35$$
Subtracting gives $$8a+32b=56\Rightarrow a+4b=7\Rightarrow a=7-4b$$
$$35=a^2-2a+b^2+14b=49-56b+16b^2-14+8b+b^2+14b=35+17b^2-34b\Rightarrow b^2-2b=0$$
$$b=0\Rightarrow a=7-4b=7; b=2 \Rightarrow a=7-4b=-1$$
$$(x-7)^2+y^2=85 \mathrm{\ or\ } (x+1)^2+(y-2)^2=85$$
