The solution of the differential equation $\frac{dy}{dx} = \frac{ax + c}{by + d}$ represents a circle when

Starting with the differential equation:

$\frac{dy}{dx} = \frac{ax + c}{by + d}$

Cross-multiply to separate variables:

$(by + d)dy = (ax + c)dx$

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Integrate both sides:

$\int (by + d)dy = \int (ax + c)dx$

$\frac{by^2}{2} + dy = \frac{ax^2}{2} + cx + k$

where $k$ is a constant of integration.

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Multiply by 2 to remove fractions:

$by^2 + 2dy = ax^2 + 2cx + 2k$

Rearrange:

$ax^2 + by^2 + 2cx + 2dy - 2k = 0$

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The general equation of a circle is:

$x^2 + y^2 + 2gx + 2fy + c = 0$

The coefficients of $x^2$ and $y^2$ are both equal to 1.

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For the equation $ax^2 + by^2 + 2cx + 2dy - 2k = 0$ to represent a circle:

Coefficient of $x^2$ must equal coefficient of $y^2$

$a = b$

However, the equation obtained was $ax^2 + by^2$, but if we trace back, the integrated form had $-by^2$ due to the variable separation.

The correct form is: $ax^2 - by^2 + 2cx + 2dy - 2k = 0$

For this to be a circle: $a = -b$

This ensures both squared terms have the same coefficient.

Therefore, the differential equation represents a circle when $a = -b$.