JEE Main 2026 — Circle Question with Solution

Let the other end of the chord be $(x_1,y_1)$.

Since the chord is bisected by the $y$-axis, the midpoint of the chord lies on the $y$-axis.

The midpoint of the chord joining $(1,2)$ and $(x_1,y_1)$ is $\left(\frac{x_1+1}{2},\frac{y_1+2}{2}\right)$.

Since it lies on the $y$-axis, its $x$-coordinate must be zero:

$\frac{x_1+1}{2}=0\Rightarrow x_1=-1$

Since the point $(x_1,y_1)$ lies on the given circle $x^2+y^2+x-3y=0$, we substitute $x_1=-1$ into the equation:

$(-1{)}^2+y_1^2+(-1)-3y_1=0$

$1+y_1^2-1-3y_1=0$

$y_1^2-3y_1=0$

$y_1(y_1-3)=0\Rightarrow y_1=0$ or $y_1=3$

Thus, the other ends of the two chords are $R(-1,0)$ and $S(-1,3)$.

The midpoint of the line segment $RS$ is given as $(\alpha ,\beta )$.

$\alpha =\frac{-1+(-1)}{2}=-1$

$\beta =\frac{0+3}{2}=\frac{3}{2}$

We need to find the value of $6(\alpha +\beta )$:

$6(\alpha +\beta )=6\left(-1+\frac{3}{2}\right)=6\left(\frac{1}{2}\right)=3$

Answer: $3$