If the solution of the differential equation

$\frac{d y}{d x}=\frac{a x+3}{2 y+f}$

represents a circle, then the value of 'a' is

-2

We have,

$\frac{d y}{d x}=\frac{a x+3}{2 y+f} \Rightarrow(a x+3) d x=(2 y+f) d y$

On integrating, we obtain

$a \frac{x^2}{2}+3 x=y^2+f y+C \Rightarrow-\frac{a}{2} x^2+y^2-3 x+f y+C=0$

This will represent a circle, if

$-\frac{a}{2}=1$              [∵ Coeff. of $x^2$ = Coeff. of $y^2$]

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

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