Question

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

• A. 2
• B. -2
• C. 3
• D. -4

Answer 1

The correct option is B -2

We have, $\frac{dy}{dx}=\frac{ax+3}{2y+f}\Rightarrow (ax+3)dx=(2y+f)dy$
On intergrating, we obtain
$a\frac{x^2}{2}+3x=y^2+fy+c\Rightarrow -\frac{a}{2}{x}^2+y^2-3x+fy+c=0$
This will represent a circle, if
$-\frac{a}{2}=1[\because Coeff.of{x}^2=Coeff.of{y}^2]and,\frac{9}{4}+\frac{f^2}{4}-c>0[Using{g}^2+\frac{f^2}{4}-c>0]\Rightarrow a=-2and9+f^2-4c>0$