Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

(N/A) The equation of a circle in the first quadrant with centre $(a, a)$ and radius $(a)$ which touches the coordinate axes is:
$(x-a)^{2}+(y-a)^{2}=a^{2}$ $(1)$
Differentiating equation $(1)$ with respect to $x$,we get:
$2(x-a)+2(y-a) y^{\prime}=0$
$(x-a)+(y-a) y^{\prime}=0$
$x-a+yy^{\prime}-ay^{\prime}=0$
$x+y y^{\prime}-a(1+y^{\prime})=0$
$a=\frac{x+y y^{\prime}}{1+y^{\prime}}$
Substituting the value of $a$ in equation $(1)$,we get:
$\left[x-\left(\frac{x+yy^{\prime}}{1+y^{\prime}}\right)\right]^{2}+\left[y-\left(\frac{x+yy^{\prime}}{1+y^{\prime}}\right)\right]^{2}=\left(\frac{x+yy^{\prime}}{1+y^{\prime}}\right)^{2}$
$\left[\frac{x+xy^{\prime}-x-yy^{\prime}}{1+y^{\prime}}\right]^{2}+\left[\frac{y+yy^{\prime}-x-yy^{\prime}}{1+y^{\prime}}\right]^{2}=\left[\frac{x+yy^{\prime}}{1+y^{\prime}}\right]^{2}$
$\left[\frac{x y^{\prime}-y y^{\prime}}{1+y^{\prime}}\right]^{2}+\left[\frac{y-x}{1+y^{\prime}}\right]^{2}=\left[\frac{x+yy^{\prime}}{1+y^{\prime}}\right]^{2}$
$(y^{\prime})^{2}(x-y)^{2}+(y-x)^{2}=(x+yy^{\prime})^{2}$
$(x-y)^{2}[1+(y^{\prime})^{2}]=(x+yy^{\prime})^{2}$