Question

A line meets the circle $x^2+y^2-4x-4y-8=0$ in two points A and B. If P(2,-2) is a point on the circle such that PA = PB = 2 then the equation of the line AB is

• A. $2x+3y=0$
• B. $3x+2y=0$
• C. $2x+3=0$
• D. $2y+3=0$

Hint:

When a point P on a circle is equidistant from the endpoints of a chord AB, the line joining the center C and P is the perpendicular bisector of the chord AB. This geometric property is often the key to solving such problems efficiently.

Answer 1

The Correct Option is D

Step 1: Analyze Geometry: Circle Centre \( C(2,2) \), Radius \( \sqrt{4+4+8} = 4 \). Point \( P(2,-2) \) is on the circle. Since \( PA=PB \), P lies on the perpendicular bisector of chord AB. Also, the centre C lies on the perpendicular bisector of any chord. So, line PC is the perpendicular bisector of AB. Line PC connects \( (2,2) \) and \( (2,-2) \), which is the vertical line \( x=2 \). Thus, the chord AB is perpendicular to PC, i.e., AB is a horizontal line \( y=k \).
Step 2: Find k: Let coordinates of A be \( (x, k) \). Given \( PA=2 \), so \( PA^2 = 4 \). \( (x-2)^2 + (k-(-2))^2 = 4 \implies (x-2)^2 + (k+2)^2 = 4 \). Also A is on circle: \( (x-2)^2 + (k-2)^2 = 16 \). Substitute \( (x-2)^2 \): \( 4 - (k+2)^2 + (k-2)^2 = 16 \) \( 4 - (k^2+4k+4) + (k^2-4k+4) = 16 \) \( 4 - 8k = 16 \) \( -8k = 12 \implies k = -3/2 \). Equation of line AB: \( y = -3/2 \implies 2y+3=0 \).