Question

If the smallest circle through the points of intersection of \( x^2 + y^2 = a^2 \) and \( x \cos \alpha + y \sin \alpha = p \), \( 0<p<a \), is \[ x^2 + y^2 - a^2 + \lambda(x \cos \alpha + y \sin \alpha - p) = 0 \] then \( \lambda = \)

• A. \( 1 \)
• B. \( -1 \)
• C. \( -p \)
• D. \( -2p \)

Hint:

To find the circle passing through points of intersection of a circle and a line, use the combined equation form: \( \text{circle} + \lambda(\text{line}) = 0 \). For the smallest such circle, \( \lambda \) is chosen to make the chord the diameter.

Answer 1

The Correct Option is D

Step 1: Understand the setup.
We are given two curves: - A circle: \( x^2 + y^2 = a^2 \) - A line: \( x \cos \alpha + y \sin \alpha = p \) These intersect at two points (since \( 0<p<a \), the line intersects the circle). The smallest circle through these two points is the circle having the segment between them as diameter. Step 2: Locus of points on a circle.
The required circle passes through the intersection points of the given circle and line. So we find their intersection, and then the locus of all circles passing through them. To find the circle passing through the points of intersection of a circle and a line, we can write its equation as: \[ x^2 + y^2 - a^2 + \lambda(x \cos \alpha + y \sin \alpha - p) = 0 \] Step 3: Use geometric conditions.
The smallest circle through two points is the one for which the line joining them is the diameter. So the center lies at the midpoint of the chord, and its radius is half the chord length. Using Lagrange multipliers or known analytic geometry result (from vector geometry), the required \( \lambda \) to minimize the radius (or area) of such a circle is: \[ \lambda = -2p \]