Question 1123419
There are two solutions (corresponding to points C and D) as shown in the drawing below.   

Line AB has the equation  {{{ y = (1/2)x - 1 }}}
Line CD has the equation {{{ y = -2x - 1 }}}

{{{ abs(AB)^2 = (4-(-4))^2 + (1-(-3))^2 = 64+16 = 80 }}}

Using point C(Cx, Cy), the distance squared to C from B  (and A) must be  80.  Using B's coordinates:
{{{ (Cx-4)^2 + (Cy-1)^2 = 80 }}}
{{{  (Cx^2-8Cx+16) + (Cy^2-2Cy+1) = 80 }}}
Subs {{{ Cy = -2Cx - 1 }}}  and  {{{Cy^2 = 4Cx^2 + 4Cx +1}}} :
 
{{{ (Cx^2-8Cx+16) + (4Cx^2+4Cx+1) + (4Cx + 2) + 1 = 80 }}}

Which resolves to   C at  (  {{{2*sqrt(3) }}} , {{{ (-4*sqrt(3))-1  }}} )        <br>

and   D at  (  {{{-2*sqrt(3) }}} , {{{ (4*sqrt(3))-1  }}} )        <br>


{{{ drawing(500,500, -10,10, -10, 10, grid(0),
       line(-4,-3, 4, 1),
       blue(circle(-4,-3,8.944)),
       green(circle(4,1, 8.944)),
       line(-3.464, 5.928, 3.464, -7.928),
        line(4,1, 3.464, -7.928),
  line(-3.464, 5.928, 4,1),
  line(-3.464, 5.928, -4,-3),
  line(-4,-3, 3.464, -7.928),
  locate(-6,-3, "A(-4,-3)"),
  locate(4.5,1.5, "B(4,1)"),

       locate(-3.6,6.5, "D"),

       locate(3.6,-7.8, "C")
) }}}