The differential equation of the family of circles passing through the origin and having centres on the x-axis is :

Options

A. `2xy "" (dy)/(dx) = x^(2) - y^(2)`
B. `2xy "" (dy)/(dx) = y^(2) - x^(2)`
C. `2xy"" (dy)/(dx) = x^(2) + y^(2)`
D. `2xy "" (dy)/(dx) + x^(2) + y^(2) = 0`

Answer

To find the differential equation of the family of circles passing through the origin and having centers on the x-axis, we can follow these steps:

### Step 1: Write the equation of the circle
The general equation of a circle with center at \((a, 0)\) and radius \(a\) is given by:
\[ (x - a)^2 + (y - 0)^2 = a^2 \]

### Step 2: Expand the equation
Expanding the equation gives:
\[ (x - a)^2 + y^2 = a^2 \]
\[ x^2 - 2ax + a^2 + y^2 = a^2 \]
Canceling \(a^2\) from both sides results in:
\[ x^2 + y^2 - 2ax = 0 \]

### Step 3: Differentiate the equation
Now, we differentiate the equation with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2ax) = 0 \]
This gives:
\[ 2x + 2y \frac{dy}{dx} - 2a = 0 \]

### Step 4: Rearrange the differentiated equation
Rearranging the equation, we have:
\[ 2y \frac{dy}{dx} = 2a - 2x \]
Dividing by 2:
\[ y \frac{dy}{dx} = a - x \]

### Step 5: Express \(a\) in terms of \(x\) and \(y\)
From the original equation \(x^2 + y^2 - 2ax = 0\), we can express \(a\) as:
\[ a = \frac{x^2 + y^2}{2x} \]

### Step 6: Substitute \(a\) back into the differentiated equation
Substituting \(a\) into \(y \frac{dy}{dx} = a - x\):
\[ y \frac{dy}{dx} = \frac{x^2 + y^2}{2x} - x \]
This simplifies to:
\[ y \frac{dy}{dx} = \frac{x^2 + y^2 - 2x^2}{2x} = \frac{y^2 - x^2}{2x} \]

### Step 7: Final form of the differential equation
Multiplying through by \(2x\) gives:
\[ 2xy \frac{dy}{dx} = y^2 - x^2 \]
Rearranging gives us the final form of the differential equation:
\[ 2xy \frac{dy}{dx} + x^2 + y^2 = 0 \]

### Conclusion
Thus, the differential equation of the family of circles passing through the origin and having centers on the x-axis is:
\[ 2xy \frac{dy}{dx} + x^2 + y^2 = 0 \]