The orthogonal trajectories of the family of circles given by `x^2 + y^2 - 2ay = 0`, is

Options

A. `x^(2)+y^(2)-2kx=0`
B. `x^(2)+y^(2)-2ky=0`
C. `x^(2)+y^(2)-2k_(1)x-2k_(2)y=0`
D. none of these

Explanation

To find the orthogonal trajectories of the family of circles given by the equation \( x^2 + y^2 - 2ay = 0 \), we will follow these steps:

### Step 1: Differentiate the given equation
The given equation is:
\[ x^2 + y^2 - 2ay = 0 \]
We will differentiate this equation with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2ay) = 0 \]
This gives us:
\[ 2x + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0 \]

### Step 2: Rearranging the equation
Rearranging the differentiated equation:
\[ 2x + (2y - 2a)\frac{dy}{dx} = 0 \]
This can be simplified to:
\[ \frac{dy}{dx} = \frac{-2x}{2y - 2a} = \frac{-x}{y - a} \]

### Step 3: Express \( a \) in terms of \( x \) and \( y \)
From the original equation, we can express \( a \):
\[ a = y - \frac{x^2 + y^2}{2y} \]
Substituting this back into our equation gives:
\[ \frac{dy}{dx} = \frac{-x}{y - (y - \frac{x^2 + y^2}{2y})} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{-x}{\frac{x^2}{2y}} = \frac{-2xy}{x^2} \]

### Step 4: Finding the orthogonal trajectories
The orthogonal trajectories will have slopes that are negative reciprocals. Therefore, we have:
\[ \frac{dy}{dx} = \frac{2y}{x} \]

### Step 5: Separating variables
We can separate variables:
\[ \frac{dy}{y} = \frac{2dx}{x} \]

### Step 6: Integrating both sides
Integrating both sides:
\[ \int \frac{dy}{y} = \int \frac{2dx}{x} \]
This gives:
\[ \ln|y| = 2\ln|x| + C \]
Exponentiating both sides results in:
\[ y = kx^2 \quad (where \, k = e^C) \]

### Step 7: Rearranging to standard form
Rearranging gives:
\[ y^2 = kx^2 \]
This represents the orthogonal trajectories of the given family of circles.

### Final Result
The orthogonal trajectories of the family of circles \( x^2 + y^2 - 2ay = 0 \) are given by:
\[ y^2 = kx^2 \]
or equivalently,
\[ x^2 + y^2 - 2ky = 0 \]