JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Circle

Let a chord of circle $C:x^2+y^2=4$ be bisected at $M(x_1,y_1)$. By $T=S_1$, its equation is $x{x}_1+y{y}_1=x_1^2+y_1^2$. Since the chord passes through $(\alpha ,0)$:

If $M$ lies on the parabola $P:y^2=8x$, then $y_1^2=8x_1$, so

One solution is $x_1=0$ (forcing $y_1=0$ too), giving the midpoint at the origin — this is the diameter along the x-axis ($y=0$), which passes through $(\alpha ,0)$ for every $\alpha$ and is always bisected at the origin (which lies on $P$). This is always one valid chord.

For $x_1\ne 0$: $\alpha =x_1+8⟹x_1=\alpha -8$, with $y_1^2=8x_1=8(\alpha -8)$, giving two more chords (with $y_1=\pm \sqrt{8(\alpha -8)}$), provided $x_1>0$ (i.e. $\alpha >8$) and the midpoint lies strictly inside the circle:

Let $u=\alpha -8$: $u^2+8u-4<0⟹u\in (-4-2\sqrt{5},-4+2\sqrt{5})$.

Combined with $u>0$ (needed for two extra real, distinct chords): $u\in (0,-4+2\sqrt{5})$, i.e.