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Maths Question 13 – JEE-MAIN 2026

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

If a chord is bisected by the y-axis, the x-coordinate of its midpoint is 0. This implies a specific relationship between the x-coordinates of the chord's endpoints.

● Video Walkthrough
Step 1: Determine the x-coordinate of the other end of the chord✦ Active

Let the given point be P(1,2). Let Q(x2,y2) be the other end of a chord drawn from P. The midpoint of PQ is M=(1+x22,2+y22). Since the chord is bisected by the y-axis, the x-coordinate of its midpoint must be 0. Therefore, 1+x22=0, which implies 1+x2=0, so x2=−1. This means both other ends, R and S, have an x-coordinate of −1.

Step 2: Find the y-coordinates of the other ends (R and S)○ Expand

Since the points R and S lie on the circle x2+y2+x−3y=0 and their x-coordinate is −1, substitute x=−1 into the circle's equation:

(−1)2+y2+(−1)−3y=0 ⟹1+y2−1−3y=0 ⟹y2−3y=0 ⟹y(y−3)=0

This gives two possible y-coordinates: y=0 or y=3. Thus, the other ends of the chords are R(−1,0) and S(−1,3).

Step 3: Calculate the midpoint of RS and the final expression○ Expand

The midpoint of the line segment RS is (α,β). Using the midpoint formula for R(−1,0) and S(−1,3):

α=−1+(−1)2=−22=−1 β=0+32=32

So, (α,β)=(−1,32). We need to find the value of 6(α+β):

6(α+β)=6(−1+32) =6(−2+32) =6(12) =3
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