We are told that the lowest part of the fence (that is, the fence at the point in the park at which it is hardest to get a home run) is to be 10 feet high. This will be at the point furthest from home plate. That point will be somewhere on the quarter circle.

Input := 

s1 = Solve[(x-250)^2 + (y-150)^2 == 175^2, y]

Output =

                                              2
       300 - Sqrt[90000 - 4 (54375 - 500 x + x )]
{{y -> ------------------------------------------}, 
                           2
 
                                               2
        300 + Sqrt[90000 - 4 (54375 - 500 x + x )]
  {y -> ------------------------------------------}}
                            2

The second solution gives us the top of the circle, and we know the furthest point from the origin lies on the top of the circle.

Input := 

ycirc[x_] = s1[[2,1,2]];

Input := 

d[x_] = Sqrt[x^2 + ycirc[x]^2];

Input := 

Plot[d[x],{x,250,425}];

Input := 

s2 = Solve[d'[x]==0,x] //N

Output =

{{x -> 400.061}}

Input := 

xfar = s2[[1,1,2]];

Input := 

ycirc[xfar]

Output =

240.037

Input := 

d[xfar]

Output =

466.548

So that's the most distant point on the outfield fence, and at that point the height of the fence should be 10 feet.