Conceptually, an easy way to obtain the solution using tangency properties:
Given
$$ \cases{ \|p-p_a\|+\frac 12\|p-p_c\|=f\\ \|p-p_b\|^2 = 4\\ p_a = (1,0)\\ p_b = (5,0)\\ p_c = (0,5)} $$
at tangency, eliminating $y$ we get
$$ p(x) = f^8-48 f^6 x+4 f^6+904 f^4 x^2-1808 f^4 x+4366 f^4-4672 f^2 x^3+1744 f^2 x^2+39088 f^2 x-159276 f^2+26896 x^4-280768 x^3+1448104 x^2-3733872 x+4756761 $$
this polynomial should have a double root (due to tangency) or
$$ p(x) = k(x-x_1)^2(x^2+a x + b)\ \ \forall x $$
or equivalently
$$ \left\{ \begin{array}{rcl} -b k x_1^2+f^8+4 f^6+4366 f^4-159276 f^2+4756761 & = & 0\\ -a k x_1^2+2 b k x_1-48 f^6-1808 f^4+39088 f^2-3733872 & = & 0\\ 2 a k x_1-b k+904 f^4+1744 f^2-k x_1^2+1448104 & = & 0\\ -a k-4672 f^2+2 k x_1-280768 & = & 0\\ 26896-k & = & 0\\ \end{array} \right. $$
now solving for $\{f,k,a,b,x_1\}$ we obtain the solution.
Follows a MATHEMATICA scrip to solve the problem.
pa = {1, 0};
pb = {5, 0};
pc = {0, 5};
p = {x, y};
c1 = Sqrt[(p - pc) . (p - pc)] + 1/2 Sqrt[(p - pa) . (p - pa)] - f
c2 = (p - pb) . (p - pb) - 2^2
polx = First[Eliminate[{c1 == 0, c2 == 0}, y]] - k (x - x1)^2 (x^2 + a x + b)
Solve[CoefficientList[polx, x] == 0, {k, f, x1, a, b}, Reals]
gr1 = ContourPlot[Sqrt[(p - pc) . (p - pc)] + 1/2 Sqrt[(p - pa) . (p - pa)] == Sqrt[41], {x, -1, 8}, {y, -3, 6}];
gr2 = ContourPlot[(p - pb) . (p - pb) - 2^2 == 0, {x, -1, 8}, {y, -3, 6}];
points = {pa, pb, pc};
gr0 = Graphics[{Red, PointSize[0.02], Point[points]}];
Show[gr1, gr2, gr0]
