Deriving differential equation satisfied by family of circles

Question

There's an exercise in Differential Equations with Applications by George Simmons which asks to derive the differential equation satisfied by the family of circles in the plane whose centers lie on the line $ y = x $ and are tangent to the $ x $ and $ y $ axes.

I've derived the general equation for this one-parameter family, $ (x - c)^2 + (y - c)^2 = c^2 $, with parameter $ c $. Using implicit differentiation, the curves also satisfy $ (x - c) + (y - c) y' = 0 $, which solves for $ c $ in terms of the other variables as $$ c = \frac{x + y y'}{1 + y'}. $$

I've tried substituting $ c $ into the first equation but I am unable to simplify the result. I feel like there is some trick or algebraic manipulation I am forgetting to arrive to the answer.

The official answer in the back of the book is $$ (x - y)^2 (1 + y'^2) = (x + y y')^2. $$ Mathematica confirms this so it's probably not an error. Can anyone nudge me in the right direction?

Answer 1

From your solution $$ c = \frac{x + y y'}{1 + y'} $$ it is straightforward to compute: $$ \begin{aligned} x-c & = y' \frac{x - y }{1 + y'} \\ y-c &= \frac{y-x }{1 + y'} \end{aligned} $$ Now substitute in the implicit equation of the circles: $$\begin{aligned} (x-c)^2 &+(y-c)^2 &=& c^2 \\ (y')^2\,\left(\frac{x - y }{1 + y'}\right)^2 &+ \left(\frac{y-x }{1 + y'}\right)^2 &=& \left(\frac{x + y y'}{1+y'}\right)^2 \end{aligned} $$ and so on...