The orthogonal trajectories of the circle `x^(2)+y^(2)-ay=0`, (where a is a parameter), is

Options

A. `x^(2)+y^(2)-ay=0`
B. `x^(2)+y^(2)=Cx`
C. `x^(2)+y^(2)=C`
D. `x^(2)+y^(2)=C(x+y)`

Explanation

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

### Step 1: Rewrite the Circle's Equation
The equation of the circle can be rewritten as:
\[ x^2 + (y - \frac{a}{2})^2 = \frac{a^2}{4} \]
This shows that the circle is centered at \( (0, \frac{a}{2}) \) with radius \( \frac{a}{2} \).

### Step 2: Differentiate the Circle's Equation
To find the slope of the tangent to the circle, we differentiate the equation \( x^2 + y^2 - ay = 0 \) with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - a\frac{dy}{dx} = 0 \]
This gives:
\[ 2x + 2y\frac{dy}{dx} - a\frac{dy}{dx} = 0 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging the equation, we have:
\[ 2y\frac{dy}{dx} - a\frac{dy}{dx} = -2x \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}(2y - a) = -2x \]
Thus,
\[ \frac{dy}{dx} = \frac{-2x}{2y - a} \]

### Step 4: Find the Slope of the Orthogonal Trajectories
The slopes of the orthogonal trajectories are negative reciprocals of the slopes of the original curves. Therefore, we have:
\[ \frac{dx}{dy} = -\frac{2y - a}{-2x} = \frac{2y - a}{2x} \]

### Step 5: Separate Variables
We can separate the variables:
\[ 2x \, dx = (2y - a) \, dy \]

### Step 6: Integrate Both Sides
Integrating both sides:
\[ \int 2x \, dx = \int (2y - a) \, dy \]
This gives:
\[ x^2 = y^2 - ay + C \]
where \( C \) is the constant of integration.

### Step 7: Rearranging the Equation
Rearranging the equation, we can write:
\[ y^2 - ay + x^2 - C = 0 \]
This represents the family of orthogonal trajectories.

### Final Form of the Orthogonal Trajectories
The orthogonal trajectories can be expressed as:
\[ y^2 + x^2 = ay + C \]