So you'll have something to solve yourself, instead of doing it for you, 
I'll do one EXACTLY LIKE IT with the numbers changed.  You solve yours
exactly like it:

Let (h,k) be the center of the circle.  Then the equation of the circle is 

(x-h)² + (y-k)² = r²

Endpoints of a chord of a circle lie on the circle.
Therefore (1,4) and (2,5) are points on the circle, and we 
can substitute them into the equation of the circle and 
have a true equation:

(1-h)² + (4-k)² = r²
(2-h)² + (5-k)² = r²

Since both equal r², we can set them equal:

(1-h)² + (4-k)² = (2-h)² + (5-k)²

1-2h+h² + 16-8k+k² = 4-4h+h² + 25-10k+k²
17-2h-8k+h²+k² = 29-4h-10k+h²+k²
      17-2h-8k = 29-4h-10k 
         2h+2k = 12
           h+k = 6

Since we are told that the line y = -4x+3 is the perpendicular bisector
of the chord EF.  We don't need to know which chord EF is, for ANY line
that is the perpendicular bisector of ANY chord must pass through the
center of the circle. 

So the center (h,k) must lie on the line y = -4x+3, and we 
can substitute them into the equation of that line and 
have a true equation:

y = -4x+3
k = -4k+3

So we have this system of equations:

h+k=6
k = -4h+3

We use substitution:

h+(-4h+3) = 6
   h-4h+3 = 6
    -3h+3 = 6
      -3h = 3
        h = -1

k = -4(-1)+3
k = 4+3
k = 7

So the center is (h,k) = (-1,7)

Use that as a model to solve yours.

Edwin