The general solution of `(dy)/(dx) = (ax +h)/(by + k)` represents a circle only when

Options

A. a = b = 0
B. a = - b `ne 0`
C. `a = b ne 0 ` , h = k
D. `a = b ne 0`

Answer

To determine when the general solution of the differential equation \(\frac{dy}{dx} = \frac{ax + h}{by + k}\) represents a circle, we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given differential equation:
\[ \frac{dy}{dx} = \frac{ax + h}{by + k} \]
We can rearrange this to facilitate integration:
\[ (by + k) dy = (ax + h) dx \]

### Step 2: Integrate Both Sides
Next, we integrate both sides:
\[ \int (by + k) dy = \int (ax + h) dx \]
The left side integrates to:
\[ \frac{by^2}{2} + ky \]
And the right side integrates to:
\[ \frac{ax^2}{2} + hx \]
Thus, we have:
\[ \frac{by^2}{2} + ky = \frac{ax^2}{2} + hx + C \]

### Step 3: Rearranging the Equation
Rearranging the equation gives us:
\[ \frac{by^2}{2} - \frac{ax^2}{2} + ky - hx - C = 0 \]
Multiplying through by 2 to eliminate the fractions, we obtain:
\[ by^2 - ax^2 + 2ky - 2hx - 2C = 0 \]

### Step 4: Compare with the General Equation of a Circle
The general equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
For our equation to represent a circle, the coefficients of \(x^2\) and \(y^2\) must be equal and non-zero.

### Step 5: Set Conditions for Coefficients
From our equation:
- The coefficient of \(y^2\) is \(b\)
- The coefficient of \(x^2\) is \(-a\)

For the equation to represent a circle, we need:
1. \(b = -a\) (so that the coefficients of \(x^2\) and \(y^2\) are equal)
2. \(a \neq 0\) and \(b \neq 0\) (to ensure the terms do not cancel out)

### Conclusion
Thus, the conditions under which the general solution of the given differential equation represents a circle are:
- \(a = -b\)
- \(a \neq 0\) and \(b \neq 0\)