We know that the equation of a circle in the first quadrant with centre (a, a) and radius a which touches the coordinate axes is :

(x -a)² + (y –a)² = a² -----------(1)

Now differentiating above equation w.r.t. x, we get,

2(x-a) + 2(y-a) = 0

⇒ (x – a) + (y – a)y’ = 0

⇒ x – a +yy’ – ay’ = 0

⇒ x + yy’ –a(1+y’) = 0

⇒ a =

Now, substituting the value of a in equation (1), we get,

⇒ (x - y)².y’² + (x – y)² = (x + yy’)²

⇒ (x – y)²[1 + (y’)²] = (x + yy’)²

Therefore, the required differential equation of the family of circles is

(x – y)²[1 + (y’)²] = (x + yy’)²