If the solution of the differential equation represents a circle, then find the value of .

Given $\frac{dy}{dx} = \frac{ax+3}{2y+5}$. Separating variables: $(2y + 5) dy = (ax + 3) dx$. Integrating both sides: $\int (2y + 5) dy = \int (ax + 3) dx \Rightarrow y^2 + 5y = \frac{a}{2}x^2 + 3x + C$. Rearranging: $-\frac{a}{2}x^2 - 3x + y^2 + 5y - C = 0$. For this quadratic equation to represent a circle, the coefficients of $x^2$ and $y^2$ must be equal and non-zero (with no xy term): $-\frac{a}{2} = 1 \Rightarrow a = -2$.