If the line `x cos alpha + y sin alpha=p` cuts the circle `x^(2)+y^(2)=a^(2)` in M and N, then the equation of the circle on MN as diameter is `x^(2)+y^(2)-a^(2)=2p(x cos alpha +y sin alpha-p)`

Answer

To solve the problem, we need to find the equation of the circle that has the line \( x \cos \alpha + y \sin \alpha = p \) as a diameter, where this line intersects the circle \( x^2 + y^2 = a^2 \) at points M and N.

### Step-by-step Solution:

1. **Identify the given equations**:
- The equation of the line is given by:
\[ x \cos \alpha + y \sin \alpha = p \]
- The equation of the circle is:
\[ x^2 + y^2 = a^2 \]

2. **Rearranging the line equation**:
- We can rearrange the line equation into standard form:
\[ x \cos \alpha + y \sin \alpha - p = 0 \]

3. **Finding the intersection points M and N**:
- To find the points of intersection of the line and the circle, we substitute \( y \) from the line equation into the circle equation.
- From the line equation, solve for \( y \):
\[ y = \frac{p - x \cos \alpha}{\sin \alpha} \]
- Substitute this expression for \( y \) into the circle equation:
\[ x^2 + \left(\frac{p - x \cos \alpha}{\sin \alpha}\right)^2 = a^2 \]
- This will yield a quadratic equation in terms of \( x \).

4. **Finding the center of the circle on MN**:
- The center of the circle that has MN as its diameter can be found using the midpoint formula. The coordinates of the midpoint \( C \) of points M and N can be derived from the intersection points.

5. **Equation of the circle with diameter MN**:
- The general equation of a circle with center \( (h, k) \) and radius \( r \) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
- Since MN is the diameter, we can express the equation of the circle in terms of the line:
\[ x^2 + y^2 - a^2 = 2p(x \cos \alpha + y \sin \alpha - p) \]
- This is the required equation of the circle on MN as diameter.

### Final Equation:
Thus, the equation of the circle on MN as diameter is:
\[ x^2 + y^2 - a^2 = 2p(x \cos \alpha + y \sin \alpha - p) \]