I would use

Manipulate[ContourPlot[(1 - b)^2*(R - r)^2 == 2*(1 + b)^2*r, {b, -1, 1}, {r, -5,5}], {R, 0.5, 2.0, 0.001}]

Then you can use the options in ContourPlot (ImageSize, PlotPoints, etc.) to make the display fit your needs. ContourPlot could be a bit slow (think of what it is doing), so each time you play with the value of R in the Manipulate scale you may need to wait a bit.

Best,

OL.

[deleted double post]

In my post above, I put the output in comment markers - that is, anything between the (* and the *) are commented out.

So when you type the Solve command (note that you use two equal signs and not one, look at the help for Solve to see examples),

Solve[((1 - b)^2)/r^2 == 2*((1 + b)^2)/(R - r)^2, r]

this is the output,

{{r -> (-R + 2 b R - b^2 R - 
    Sqrt[2] Sqrt[R^2 - 2 b^2 R^2 + b^4 R^2])/(
   1 + 6 b + b^2)}, {r -> (-R + 2 b R - b^2 R + 
    Sqrt[2] Sqrt[R^2 - 2 b^2 R^2 + b^4 R^2])/(1 + 6 b + b^2)}}

This is in the form of a replacement rule ( type r^2 /. r->7 to see how a replacement rule works). But I wanted to grab the actual answers without copying and pasting them.

Here is another way to visualize your function:

define a function to be the solutions to the equations,

func[b_, R_] = 
  r /. Solve[((1 - b)^2)/r^2 == 2*((1 + b)^2)/(R - r)^2, r];
Plot[func[b, 1.0], {b, 0, 5}, Evaluated -> True]

These are the solutions for the equation when R=1.0, and you can plot them for any value of R.