JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Circle

Let $M(h,k)$ be the midpoint of a chord of the circle $x^2+(y-1{)}^2=1$ (centre $C(0,1)$) drawn from the origin $O(0,0)$.

Since $M$ is the midpoint of a chord, $CM$ is perpendicular to that chord; and the chord passes through $O$, so $OM$ lies along the chord. Hence $CM\perp OM$, which means $M$ lies on the circle having $OC$ as diameter.

Circle on diameter $OC$, with $O(0,0)$ and $C(0,1)$: centre $\left(0,\frac{1}{2}\right)$, radius $\frac{1}{2}$:


Intersect with $x+y=1$, i.e. $x=1-y$:


$y=1\Rightarrow x=0$: $P(0,1)$. $y=\frac{1}{2}\Rightarrow x=\frac{1}{2}$: $Q\left(\frac{1}{2},\frac{1}{2}\right)$.