There may me more elegant solutions, but this works:
YO need to determine the length of side b. This is determined in two steps: First, find the lengths of the other sides in terms of b; then set up an equation for b.
We are given b-c = Δ, so c is simply given as:
c = b – Δ 
To determine a, note that the area of the triangle is given by one half the base times the height and also by one half the perimeter times the inradius. To see how the latter is derived, consider a triangle in which the inscribed circle has been constructed. Draw line segments form the center of this circle to the vertices of the triangle. This divides the area of the original triangle into three components that are each one half of one side times the inradius.
Thus we have:
a= Δ + b((h_b)/r-2) 
Note that (h_b)/r must be greater than 2 because of the triangle inequality. Attempting to solve the system with (h_b)/r < 2 would still give a triangle, but it would be turned “inside out” with the “inscribed circle” lying outside the triangle; the circle would be tangent to side a and the extensions of sides b and c rather than all three sides proper.
Now equate the area as given by one half the base times the height with the area as given by Heron’s Formula:
Substitute  and  into , square both sides, clear fractions and divide by a common factor of b2 that appears (b is of course nonzero). This leads to a quadratic equation (which becomes linear in the specific case hb/r = 4:
Equation  has a unique positive root when 2 < (h_b)/r < 4, and this will give a unique triangle meeting the given values of Δ, r and (h_b) . For (h_b)/r > 4, we are not guaranteed a solution and with (h_b)/r > 4 there may be two roots.