The equation of the circle and its chord are-respectively `x^2 + y^2 = a^2 and x cos alpha + y sin alpha = p`. The equation of the circle of which this chord isa diameter is

Options

A. `x^(2)+y^(2) -px cos alpha -2py sin alpha + 2p^(2) -a^(2) =0`
B. `x^(2) +p^(2) -2px cos alpha - 2p y sin alpha + p^(2) -a^(2) =0`
C. `x^(2) + y^(2) +2px cos alpha + 2py sin alpha + 2p^(2) -a^(2) =0`
D. None of these

Answer

To find the equation of the circle for which the given chord is a diameter, we can follow these steps:

### Step 1: Write the equations of the circle and the chord
The equation of the circle is given by:
\[ x^2 + y^2 = a^2 \]
The equation of the chord is given by:
\[ x \cos \alpha + y \sin \alpha = p \]

### Step 2: Write the general equation of a circle passing through the intersection points
The general equation of any circle that passes through the intersection points of the given circle and the chord can be expressed as:
\[ x^2 + y^2 - a^2 + \lambda (x \cos \alpha + y \sin \alpha - p) = 0 \]
where \(\lambda\) is a parameter.

### Step 3: Identify the center of the circle
The center of this circle can be found by rearranging the equation:
\[ \text{Center} = \left(-\frac{\lambda \cos \alpha}{2}, -\frac{\lambda \sin \alpha}{2}\right) \]

### Step 4: Set the condition for the chord to be a diameter
For the chord to be a diameter of the circle, the center must lie on the chord. Therefore, we substitute the center coordinates into the chord equation:
\[ -\frac{\lambda \cos \alpha}{2} \cos \alpha - \frac{\lambda \sin \alpha}{2} \sin \alpha = p \]
This simplifies to:
\[ -\frac{\lambda}{2}(\cos^2 \alpha + \sin^2 \alpha) = p \]
Using the identity \(\cos^2 \alpha + \sin^2 \alpha = 1\), we have:
\[ -\frac{\lambda}{2} = p \]
Thus, we find:
\[ \lambda = -2p \]

### Step 5: Substitute \(\lambda\) back into the circle equation
Now, substituting \(\lambda = -2p\) back into the general equation of the circle gives:
\[ x^2 + y^2 - a^2 - 2p(x \cos \alpha + y \sin \alpha - p) = 0 \]
Expanding this, we get:
\[ x^2 + y^2 - a^2 - 2p x \cos \alpha - 2p y \sin \alpha + 2p^2 = 0 \]

### Step 6: Rearranging the equation
Rearranging the terms, we obtain the final equation of the circle:
\[ x^2 + y^2 - 2p x \cos \alpha - 2p y \sin \alpha + (2p^2 - a^2) = 0 \]

### Final Answer
Thus, the equation of the circle for which the given chord is a diameter is:
\[ x^2 + y^2 - 2p(x \cos \alpha + y \sin \alpha) + (2p^2 - a^2) = 0 \]

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