If the solution of the differential equation $\frac{dy}{dx}=\frac{ax+3}{2y+f}$represents a circle, then the value of 'a' is

We have,

$\begin{array}{l}\frac{dy}{dx}=\frac{ax+3}{2y+f}\Rightarrow (ax+3)dx=(2y+f)dy\\ \Rightarrow a\frac{x^2}{2}+3x=y^2+fy+C\text{ (Integrating) }\\ \Rightarrow -\frac{a}{2}{x}^2+y^2-3x+fy+C=0\end{array}$

This will represent a circle, if

$-\frac{a}{2}=1$ $\left[\because \text{ Coeff. of }{x}^2=\text{ Coeff. of }{y}^2\right]$

and, $\frac{9}{4}+\frac{{\mathrm{f}}^2}{4}-\mathrm{C}>0\left[\mathrm{Using}:{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}>0\right]$

$\Rightarrow a=-2\text{ and }9+f^2-4C>0$