Let the locus of the midpoints of the chords of the circle x²+(y-1)²=1 drawn from the origin intersect the line x+y=1 at P and Q. Then, the length of PQ is :

Solution

Maths sir · 3 steps

The centre-to-midpoint line is perpendicular to the chord, so the locus is the circle on OC as diameter.

1. Right angle at M. The circle has centre C(0,1) and radius 1, and the origin O lies on it. If M is the midpoint of a chord through O, then CM is perpendicular to the chord. So ∠ OMC = 90° and M lies on the circle with OC as diameter.

   x(x-0) + (y-0)(y-1) = 0 ⇒ x² + y² - y = 0

2. Meet the line. Put y = 1 - x into the locus: x² + (1-x)² - (1-x) = 0, which gives 2x² - x = 0. So x = 0 or x = 1/2, and the points are P(0,1) and Q(1/2,1/2).
3. So. PQ = √((1/2)² + (1/2)²) = √(1/2) = 1/√2, which is option (c).

The idea to keep. For midpoints of chords through a fixed point A, the line from the centre C to the midpoint is perpendicular to the chord. The locus is the circle with AC as diameter.

Answer from the NTA final answer key. Solution by PrepWiser.