The differential equation representing all possible curves that cut each member of the family of circles `x^(2)+y^(2)-2Cx=0` (C is a parameter) at right angle, is

Options

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

Explanation

To solve the problem, we need to find the differential equation representing all possible curves that cut each member of the family of circles given by the equation \( x^2 + y^2 - 2Cx = 0 \) at right angles.

### Step-by-Step Solution:

1. **Identify the Family of Curves**:
The given family of curves is \( x^2 + y^2 - 2Cx = 0 \). We can rewrite this as:
\[ y^2 = 2Cx - x^2 \]
This represents a family of circles with center at \( (C, 0) \) and radius \( C \).

2. **Differentiate the Equation**:
We differentiate the equation \( x^2 + y^2 - 2Cx = 0 \) with respect to \( x \):
\[ 2x + 2y \frac{dy}{dx} - 2C = 0 \]
Rearranging gives:
\[ 2y \frac{dy}{dx} = 2C - 2x \]
Simplifying further, we have:
\[ y \frac{dy}{dx} = C - x \]

3. **Express \( C \) in Terms of \( x \) and \( y \)**:
From the equation \( y \frac{dy}{dx} = C - x \), we can express \( C \) as:
\[ C = y \frac{dy}{dx} + x \]

4. **Substitute \( C \) Back into the Original Equation**:
Substitute \( C \) back into the original equation \( x^2 + y^2 - 2Cx = 0 \):
\[ x^2 + y^2 - 2(y \frac{dy}{dx} + x)x = 0 \]
This simplifies to:
\[ x^2 + y^2 - 2xy \frac{dy}{dx} - 2x^2 = 0 \]
Rearranging gives:
\[ -x^2 + y^2 - 2xy \frac{dy}{dx} = 0 \]

5. **Find the Differential Equation at Right Angles**:
For curves that intersect at right angles, the product of their slopes is \(-1\). Therefore, if \( \frac{dy}{dx} \) is the slope of the curve we are looking for, we replace \( \frac{dy}{dx} \) with \(-\frac{dx}{dy}\):
\[ -x^2 + y^2 - 2xy \left(-\frac{dx}{dy}\right) = 0 \]
This leads to:
\[ -x^2 + y^2 + 2xy \frac{dx}{dy} = 0 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 2xy \frac{dx}{dy} = x^2 - y^2 \]
Dividing through by \( 2xy \):
\[ \frac{dx}{dy} = \frac{x^2 - y^2}{2xy} \]

### Final Result:
The differential equation representing all possible curves that cut the given family of circles at right angles is:
\[ \frac{dy}{dx} = \frac{2xy}{x^2 - y^2} \]