Question

Problem: From a point \( P \) on the circle \( x^2 + y^2 = 4 \), two tangents are drawn to the circle \( x^2 + y^2 - 6x - 6y + 14 = 0 \). If \( A \) and \( B \) are the points of contact of those lines, then the locus of the center of the circle passing through the points \( P \), \( A \), and \( B \) is: Identify the correct option from the following:

• A. \( x^2 + y^2 - 3x - 3y + 4 = 0 \)
• B. \( 2x^2 + 2y^2 + 6x + 6y - 7 = 0 \)
• C. \( x^2 + y^2 + 3x + 3y - 4 = 0 \)
• D. \( 2x^2 + 2y^2 - 6x - 6y + 7 = 0 \)

Hint:

The chord of contact from a point to a circle helps in finding the locus of centers of circles passing through the point and the points of tangency.

Answer 1

The Correct Option is D

Step 1: Analyze the given circles
The first circle is \( x^2 + y^2 = 4 \), with center \( (0, 0) \) and radius \( 2 \). Point \( P (x_1, y_1) \) satisfies \( x_1^2 + y_1^2 = 4 \). The second circle is \( x^2 + y^2 - 6x - 6y + 14 = 0 \), which simplifies to \( (x - 3)^2 + (y - 3)^2 = 4 \), with center \( (3, 3) \) and radius \( 2 \). Step 2: Find the chord of contact
The chord of contact \( AB \) from \( P (x_1, y_1) \) to the second circle is given by: \[ (x - 3)(x_1 - 3) + (y - 3)(y_1 - 3) = 4 \] Simplifying: \[ x x_1 + y y_1 - 3x - 3y - 3x_1 - 3y_1 + 10 = 0 \] Step 3: Determine the locus of the center
The center of the circle passing through \( P \), \( A \), and \( B \) lies on the perpendicular from the center of the second circle to the chord of contact \( AB \), and must account for \( P \)'s position. After geometric analysis (or testing options), the locus is found to be: \[ 2x^2 + 2y^2 - 6x - 6y + 7 = 0 \] This matches option (4), which has center \( \left(\frac{3}{2}, \frac{3}{2}\right) \) and radius \( 1 \), a plausible locus for the center of the circle through \( P \), \( A \), and \( B \).