Question

The straight line \(x \cos \alpha + y \sin \alpha = p\) cuts the circle \(x^2 + y^2 - a^2 = 0\) at A and B. Then the equation of circle having AB as diameter is

• A. \(x^2 + y^2 - a^2 + p(x \cos \alpha + y \sin \alpha - p) = 0\)
• B. \(x^2 + y^2 - a^2 - p(x \cos \alpha + y \sin \alpha + p) = 0\)
• C. \(x^2 + y^2 - a^2 + 2p(x \cos \alpha + y \sin \alpha - p) = 0\)
• D. \(x^2 + y^2 - a^2 - 2p(x \cos \alpha + y \sin \alpha - p) = 0\)

Hint:

For circle with AB as diameter, use the property \((\text{Point}-A)\cdot(\text{Point}-B) = 0\) or substitute chord equation into circle and adjust coefficient to get diameter circle.

Answer 1

The Correct Option is D

Step 1: Use circle property.
If AB is a chord of circle \(x^2 + y^2 - a^2 = 0\), the circle with AB as diameter passes through A and B.

Step 2: Midpoint form.
Equation of circle with diameter AB: \((x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0\)

Step 3: Relation using line equation.
For line \(x \cos \alpha + y \sin \alpha = p\), the chord AB is perpendicular to radius through midpoint, and we derive formula:
\[ x^2 + y^2 - a^2 - 2p(x \cos \alpha + y \sin \alpha - p) = 0 \]

Step 4: Final conclusion.
Hence, the required circle equation is \[ \boxed{x^2 + y^2 - a^2 - 2p(x \cos \alpha + y \sin \alpha - p) = 0} \]