Simplifying equations is a skill that can take time to learn, but it's a very vital one. One thing to note though, is that there's nothing intrinsically special about "these type of equations." If you're familiar with algebra and moving numbers about it's not too difficult to master, with some practice.
For now, let's start at the beginning... we'll begin with the distance formula. The circle graphed by the set of all points 25 units away from (8,-1) is:
\(\displaystyle \sqrt{\left(8-x\right)^2+\left(-1-y\right)^2}=25\)
After expanding the squares we're left with:
\(\displaystyle \sqrt{64-16x+x^2+1+2y+y^2}=25\)
The problem asks you to find the point(s) on this circle whose x-coordinate is 1. So we can sub in 1 for X:
\(\displaystyle \sqrt{64-16(1)+1^2+1+2y+y^2}=25\)
\(\displaystyle \sqrt{50+2y+y^2}=25\)
To eliminate the square root, simply square both sides:
\(\displaystyle 50+2y+y^2=625\)
Subtract 625 from both sides and reorder the terms:
\(\displaystyle y^2+2y-575=0\)
Now factor that. I personally prefer to use the quadratic formula, but you may have other methods that are easier for you. I'll demonstrate the quadratic formula:
\(\displaystyle y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)
\(\displaystyle y=\frac{-2\pm \sqrt{2^2-4\left(1\right)\left(-575\right)}}{2\left(1\right)}\)
\(\displaystyle y=\frac{-2\pm \sqrt{4-\left(-2300\right)}}{2}\)
\(\displaystyle y=\frac{-2\pm \sqrt{2304}}{2}\)
\(\displaystyle y=\frac{-2\pm 48}{2}\)
\(\displaystyle y=-1\pm 24\)
Hope that helps.
