The differential equation of family of curves `x^(2)+y^(2)-2ax=0`, is

Options

A. `x^(2)-y^(2)-2xyy'=0`
B. `y^(2)-x^(2)=2xyy'`
C. `x^(2)+y^(2)+2y''=0`
D. none of these

Answer

To find the differential equation of the family of curves given by the equation \( x^2 + y^2 - 2ax = 0 \), we will follow these steps:

### Step 1: Rewrite the given equation
The given equation is:
\[ x^2 + y^2 - 2ax = 0 \]
This can be rearranged to express \( a \):
\[ 2ax = x^2 + y^2 \implies a = \frac{x^2 + y^2}{2x} \]

### Step 2: Differentiate the equation
Next, we differentiate the original equation with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2ax) = 0 \]
Using the chain rule for \( y^2 \) and the product rule for \( 2ax \):
\[ 2x + 2y \frac{dy}{dx} - 2a - 2x \frac{da}{dx} = 0 \]

### Step 3: Substitute \( a \) and simplify
From our previous step, we have \( a = \frac{x^2 + y^2}{2x} \). We differentiate \( a \) with respect to \( x \):
\[ \frac{da}{dx} = \frac{(2x)(2x) - (x^2 + y^2)(2)}{(2x)^2} = \frac{4x^2 - 2(x^2 + y^2)}{4x^2} = \frac{2x^2 - 2y^2}{4x^2} = \frac{x^2 - y^2}{2x^2} \]
Now substituting \( a \) and \( \frac{da}{dx} \) back into the differentiated equation:
\[ 2x + 2y \frac{dy}{dx} - 2\left(\frac{x^2 + y^2}{2x}\right) - 2x\left(\frac{x^2 - y^2}{2x^2}\right) = 0 \]
This simplifies to:
\[ 2x + 2y \frac{dy}{dx} - \frac{x^2 + y^2}{x} - \frac{x^2 - y^2}{x} = 0 \]
Combining terms:
\[ 2y \frac{dy}{dx} = \frac{y^2 - x^2}{x} \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ y^2 - x^2 = 2xy \frac{dy}{dx} \]

### Final Result
Thus, the differential equation of the family of curves is:
\[ y^2 - x^2 = 2xy \frac{dy}{dx} \]