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Q.
If the solution of the differential equation $\frac{dy}{dx}=\frac{ax+3}{2y+f}$ represents a circle, then the value of $'a'$ is
Differential Equations
Solution:
We have, $\frac{dy}{dx}=\frac{ax+3}{2y+f}$
$\Rightarrow \left(ax+3\right)dx=\left(2y+f\right)dy$
$\Rightarrow a \frac{x^{2}}{2}+3x=y^{2}+fy+C\quad$ (Integrating)
$\Rightarrow -\frac{a}{2}x^{2}+y^{2}-3x+fy+C=0$
This will represent a circle, if $\frac{-a}{2}=1$
$\Rightarrow a=-2$
$[ \because$ coefficient of $x^2 =$ coefficient of $y^2]$