Question

The locus of midpoints of the chords of the circle \[ x^2+y^2-2x-2y+1=0 \] which are parallel to the line \[ x+y+2=0 \] is

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

Hint:

The line joining the centre of a circle to the midpoint of a chord is always perpendicular to the chord. Hence, the locus of the midpoints of a family of parallel chords is a diameter perpendicular to those chords.

Answer 1

The Correct Option is D

Step 1: Find the centre of the circle. The given circle is \[ x^2+y^2-2x-2y+1=0. \] Comparing with the standard form, \[ (x-1)^2+(y-1)^2=1, \] the centre is \[ C(1,1). \]

Step 2: Use the midpoint property. The midpoint of any chord of a circle lies on the perpendicular from the centre to the chord. The given chords are parallel to \[ x+y+2=0, \] whose slope is \[ -1. \] Hence the perpendicular has slope \[ 1. \]

Step 3: Find the required locus. The line of slope \(1\) passing through \[ (1,1) \] is \[ y-1=x-1, \] or \[ x-y=0. \] Therefore, \[ \boxed{x-y=0}. \] Thus, \[ \boxed{(D)} \] is the correct answer.