Q. If the differential equation representing the family of all circles touching $x$-axis at the origin is $\left(x^{2}-y^{2}\right) \frac{dy}{dx} = g\left(x\right) \,y,$ then $g(x)$ equals :
Solution:
$x^{2}+\left(y-a\right)^{2}=a^{2}$
$x^{2}+y^{2}-2ay=0\,...\left(i\right)$
diff. $w.r.t. x$
$2x+2y \frac{dy}{dx}-2a \frac{dy}{dx}=0$
$a=\frac{x+y.y'}{y'}\,...\left(ii\right)$
put $\left(ii\right)$ in $\left(i\right)$
$x^{2}+y^{2}-2y\left(\frac{xy.y'}{y'}\right)=0$
$\left(x^{2}-y^{2}\right)y'=2xy\,...\left(iii\right)$
compare $\left(iii\right)$ with $\left(x^{2}-y^{2}\right) \frac{dy}{dx}=g\left(x\right).y$
gives $g\left(x\right) = 2x$
