Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.

(N/A) Let $C$ denote the family of circles in the second quadrant and touching the coordinate axes. Let $(-a, a)$ be the coordinate of the centre of any member of this family.
The equation representing the family $C$ is
$(x+a)^{2}+(y-a)^{2}=a^{2}$ ............$(1)$
or $x^{2}+y^{2}+2ax-2ay+a^{2}=0$ .............. $(2)$
Differentiating equation $(2)$ with respect to $x$,we get
$2x+2y \frac{dy}{dx}+2a-2a \frac{dy}{dx} = 0$
or $x+y \frac{dy}{dx} = a \left(\frac{dy}{dx}-1\right)$
or $a = \frac{x+y y^{\prime}}{y^{\prime}-1}$
Substituting the value of $a$ in equation $(1)$,we get
$\left[x+\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^{2}+\left[y-\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^{2}=\left[\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^{2}$
or $\left[\frac{x y^{\prime}-x+x+y y^{\prime}}{y^{\prime}-1}\right]^{2}+\left[\frac{y y^{\prime}-y-x-y y^{\prime}}{y^{\prime}-1}\right]^{2}=\left[\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^{2}$
or $(x y^{\prime}+y y^{\prime})^{2}+(-y-x)^{2}=(x+y y^{\prime})^{2}$
or $(x+y)^{2} (y^{\prime})^{2}+(x+y)^{2}=(x+y y^{\prime})^{2}$
or $(x+y)^{2} [1+(y^{\prime})^{2}]=(x+y y^{\prime})^{2}$
This is the required differential equation.