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JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 13

Suppose that two chords, drawn from the point (1,2)(1,2) on the circle x2+y2+x−3y=0x^2+y^2+x-3y=0 are bisected by the yy-axis. If the other ends of these chords are R and S, and the mid point of the line segment RS is (α,β)(\alpha,\beta), then 6(α+β)6(\alpha+\beta) is equal to:

Choose one option
Answer and worked solution

Official answer: B

Other end. A chord from (1,2)(1,2) bisected at (0,m)(0,m) ends at (−1, 2m−2)(-1,\,2m-2).

On the circle. 1+(2m−2)2−1−3(2m−2)=01+(2m-2)^2-1-3(2m-2)=0, so 2m−2=02m-2=0 or 33: R=(−1,0)R=(-1,0) and S=(−1,3)S=(-1,3).

Midpoint (−1,32)\left(-1,\frac32\right), so 6(α+β)=36(\alpha+\beta)=3.

(B)

Topics:Equation of a circle
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