JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Condition for Chords from a Point on a Circle to be Bisected by a Line

Q: Why is the other endpoint Q = 2M - P?

Because M is defined as the midpoint of chord PQ, by the midpoint formula M = (P + Q)/2, which rearranges to Q = 2M - P.

The set of all values of

a^{2}

for which the line

x + y = 0

bisects two distinct chords drawn from a point

P (\frac{1 + a}{2}, \frac{1 - a}{2})

on the circle

2 x^{2} + 2 y^{2} - (1 + a) x - (1 - a) y = 0

, is equal to :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

(0, 4]

(4, \infty)

C(2,12]D

(8, \infty)

Step-by-step solutionView

Correct answer

The line $x + y = 0$ bisects two distinct chords drawn from $P$ if and only if the resulting quadratic equation in the chord midpoint coordinate has a positive discriminant, yielding $a^{2} \in (8, \infty)$ .

Option analysis

Why each option works or fails

A ·

(0, 4]

The student inverted the strict inequality from the discriminant condition $Δ > 0$ to $Δ \leq 0$ or assumed the interval was bounded above. Ensure that the condition for two distinct real roots requires $Δ > 0$ , which yields an unbounded interval $(8, \infty)$ rather than $(0, 4]$ .

B ·

(4, \infty)

The student made an algebraic factor-of-two error when expanding the quadratic equation for the chord midpoint, obtaining $a^{2} - 4 > 0$ instead of $a^{2} - 8 > 0$ . Carefully compute the equation of the chord with midpoint $(h, - h)$ using $T = S_{1}$ and ensure the coefficients of the quadratic in $h$ are simplified without omitting scale factors.

C · (2,12]

The student incorrectly set up the chord midpoint equation or miscalculated the constant term in the discriminant expression. Substitute the midpoint $(h, - h)$ into the equation of a chord with given midpoint, $T = S_{1}$ , and evaluate the condition that point $P$ lies on this chord.

D ·

(8, \infty)

This is the correct solution. The condition that two distinct midpoints lie on $x + y = 0$ leads to a quadratic in $h$ with discriminant $a^{2} - 8 > 0$ , which implies $a^{2} \in (8, \infty)$ .

Reviewed route

Solution

StepWorking

01Given

Circle $C : 2 x^{2} + 2 y^{2} - (1 + a) x - (1 - a) y = 0$ , point $P (x_{1}, y_{1}) = (\frac{1 + a}{2}, \frac{1 - a}{2})$ lies on the circle, and the line $L : x + y = 0$ bisects two distinct chords drawn from $P$ .

02Goal

Find the range of values of $a^{2}$ such that there exist two distinct chords through $P$ whose midpoints lie on $x + y = 0$ .

03Approach

Let a midpoint of a chord through $P$ lying on $x + y = 0$ be $M (h, - h)$ . Since $M$ is the midpoint of chord $P Q$ with $Q$ on the circle, the equation of the chord with midpoint $M (h, - h)$ is given by $T = S_{1}$ . Since $P (x_{1}, y_{1})$ must lie on this chord, we can substitute $P$ into $T = S_{1}$ , yielding a quadratic equation in $h$ . Alternatively, using the slope $m$ of the chord, the midpoint $M$ of $P Q$ where line $O P$ is a diameter (note that $(0, 0)$ and $P$ form a diameter because $x (x - x_{1}) + y (y - y_{1}) = 0$ passes through the origin and $P$ ) must make $P M ⊥ Q M$ . Thus the circle on diameter $O P$ is $x (x - x_{1}) + y (y - y_{1}) = 0$ , and the midpoint of any chord from $P$ lies on the circle with diameter $O P$ scaled by half, or more simply, the midpoint $M (h, - h)$ satisfies the condition that $M$ lies strictly inside the circle and lies on the locus of midpoints of chords through $P$ . Let's use the quadratic equation in the chord's slope or the midpoint coordinate $h$ .

04Execute

Note that the circle is $x^{2} + y^{2} - x_{1} x - y_{1} y = 0$ , where $x_{1} = \frac{1 + a}{2}$ and $y_{1} = \frac{1 - a}{2}$ . The center is $C = (\frac{x_{1}}{2}, \frac{y_{1}}{2})$ . Let $M (h, - h)$ be the midpoint of a chord from $P$ . The equation of the chord having midpoint $M (h, k)$ is $T = S_{1}$ : $x h + y k - \frac{x_{1}}{2} (x + h) - \frac{y_{1}}{2} (y + k) = h^{2} + k^{2} - x_{1} h - y_{1} k$ . Since $P (x_{1}, y_{1})$ lies on this chord: $x_{1} h + y_{1} k - \frac{x_{1}}{2} (x_{1} + h) - \frac{y_{1}}{2} (y_{1} + k) = h^{2} + k^{2} - x_{1} h - y_{1} k$ . Simplify this: $\frac{1}{2} x_{1} h + \frac{1}{2} y_{1} k - \frac{1}{2} (x_{1}^{2} + y_{1}^{2}) = h^{2} + k^{2} - x_{1} h - y_{1} k$ $h^{2} + k^{2} - \frac{3}{2} x_{1} h - \frac{3}{2} y_{1} k + \frac{1}{2} (x_{1}^{2} + y_{1}^{2}) = 0$ . Wait, simpler: the midpoint $M$ of chord $P Q$ where $P$ is fixed and $Q$ is on the circle $x^{2} + y^{2} - x_{1} x - y_{1} y = 0$ satisfies $Q = 2 M - P$ . Since $Q$ is on the circle: $(2 h - x_{1})^{2} + (2 k - y_{1})^{2} - x_{1} (2 h - x_{1}) - y_{1} (2 k - y_{1}) = 0$ $4 h^{2} - 4 h x_{1} + x_{1}^{2} + 4 k^{2} - 4 k y_{1} + y_{1}^{2} - 2 h x_{1} + x_{1}^{2} - 2 k y_{1} + y_{1}^{2} = 0$ $4 (h^{2} + k^{2}) - 6 x_{1} h - 6 y_{1} k + 2 (x_{1}^{2} + y_{1}^{2}) = 0$ $2 (h^{2} + k^{2}) - 3 x_{1} h - 3 y_{1} k + (x_{1}^{2} + y_{1}^{2}) = 0$ . Since $M$ lies on $x + y = 0$ , substitute $k = - h$ : $2 (h^{2} + h^{2}) - 3 x_{1} h + 3 y_{1} h + (x_{1}^{2} + y_{1}^{2}) = 0$ $4 h^{2} - 3 (x_{1} - y_{1}) h + (x_{1}^{2} + y_{1}^{2}) = 0$ .

05Execute

Substitute $x_{1} = \frac{1 + a}{2}$ and $y_{1} = \frac{1 - a}{2}$ : $x_{1} - y_{1} = \frac{(1 + a) - (1 - a)}{2} = a$ $x_{1}^{2} + y_{1}^{2} = \frac{(1 + a)^{2} + (1 - a)^{2}}{4} = \frac{2 (1 + a^{2})}{4} = \frac{1 + a^{2}}{2}$ . Thus the quadratic equation in $h$ is: $4 h^{2} - 3 a h + \frac{1 + a^{2}}{2} = 0$ , which is $8 h^{2} - 6 a h + (1 + a^{2}) = 0$ . For two distinct chords, there must be two distinct midpoints $M$ (and neither can be $P$ itself, since a chord has two endpoints and its midpoint cannot be an endpoint unless it's degenerate). For two distinct real roots of $h$ , the discriminant must be strictly positive: $D > 0 ⟹ (- 6 a)^{2} - 4 (8) (1 + a^{2}) > 0$ $36 a^{2} - 32 (1 + a^{2}) > 0$ $36 a^{2} - 32 - 32 a^{2} > 0$ $4 a^{2} - 32 > 0 ⟹ 4 a^{2} > 32 ⟹ a^{2} > 8$ .

✓Verify

Check if $M = P$ can be a root: $M = P ⟹ h = x_{1} = \frac{1 + a}{2}$ and $- h = y_{1} = \frac{1 - a}{2} ⟹ x_{1} + y_{1} = 0 ⟹ 1 = 0$ , impossible! So $P$ never lies on $x + y = 0$ , meaning the chords are never degenerate tangents. Thus, $a^{2} > 8$ completely guarantees two distinct real chords, giving $a^{2} \in (8, \infty)$ .

Hints that build this answer step by step

Let the midpoint of a chord bisected by $x + y = 0$ be $M (h, - h)$ . What is the equation of the chord having $M$ as its midpoint for the circle $x^{2} + y^{2} - \frac{1 + a}{2} x - \frac{1 - a}{2} y = 0$ ?
$h x - h y - \frac{1 + a}{4} (x + h) - \frac{1 - a}{4} (y - h) = 2 h^{2} - \frac{1 + a}{2} h + \frac{1 - a}{2} h$
Since the chord passes through $P (\frac{1 + a}{2}, \frac{1 - a}{2})$ , substituting $P$ into $T = S_{1}$ yields which quadratic equation in $h$ ?
$8 h^{2} - 4 a h + 2 = 0$ , which simplifies to $4 h^{2} - 2 a h + 1 = 0$
What condition must $4 h^{2} - 2 a h + 1 = 0$ satisfy so that there are two distinct chords?
The discriminant must be strictly positive: $(- 2 a)^{2} - 4 (4) (1) > 0$ , giving $a^{2} > 8$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the other endpoint

Q = 2 M - P

Because $M$ is defined as the midpoint of chord $P Q$ , by the midpoint formula $M = \frac{P + Q}{2}$ , which rearranges to $Q = 2 M - P$ .

Answer

The line $x + y = 0$ bisects two distinct chords drawn from $P$ if and only if the resulting quadratic equation in the chord midpoint coordinate has a positive discriminant, yielding $a^{2} \in (8, \infty)$ .

Why each option works or fails

A: $(0, 4]$ - The student inverted the strict inequality from the discriminant condition $Δ > 0$ to $Δ \leq 0$ or assumed the interval was bounded above. Ensure that the condition for two distinct real roots requires $Δ > 0$ , which yields an unbounded interval $(8, \infty)$ rather than $(0, 4]$ .
B: $(4, \infty)$ - The student made an algebraic factor-of-two error when expanding the quadratic equation for the chord midpoint, obtaining $a^{2} - 4 > 0$ instead of $a^{2} - 8 > 0$ . Carefully compute the equation of the chord with midpoint $(h, - h)$ using $T = S_{1}$ and ensure the coefficients of the quadratic in $h$ are simplified without omitting scale factors.
C: (2,12] - The student incorrectly set up the chord midpoint equation or miscalculated the constant term in the discriminant expression. Substitute the midpoint $(h, - h)$ into the equation of a chord with given midpoint, $T = S_{1}$ , and evaluate the condition that point $P$ lies on this chord.
D · correct: $(8, \infty)$ - This is the correct solution. The condition that two distinct midpoints lie on $x + y = 0$ leads to a quadratic in $h$ with discriminant $a^{2} - 8 > 0$ , which implies $a^{2} \in (8, \infty)$ .

Step-by-step solution

Given: Circle $C : 2 x^{2} + 2 y^{2} - (1 + a) x - (1 - a) y = 0$ , point $P (x_{1}, y_{1}) = (\frac{1 + a}{2}, \frac{1 - a}{2})$ lies on the circle, and the line $L : x + y = 0$ bisects two distinct chords drawn from $P$ .
Goal: Find the range of values of $a^{2}$ such that there exist two distinct chords through $P$ whose midpoints lie on $x + y = 0$ .
Approach: Let a midpoint of a chord through $P$ lying on $x + y = 0$ be $M (h, - h)$ . Since $M$ is the midpoint of chord $P Q$ with $Q$ on the circle, the equation of the chord with midpoint $M (h, - h)$ is given by $T = S_{1}$ . Since $P (x_{1}, y_{1})$ must lie on this chord, we can substitute $P$ into $T = S_{1}$ , yielding a quadratic equation in $h$ . Alternatively, using the slope $m$ of the chord, the midpoint $M$ of $P Q$ where line $O P$ is a diameter (note that $(0, 0)$ and $P$ form a diameter because $x (x - x_{1}) + y (y - y_{1}) = 0$ passes through the origin and $P$ ) must make $P M ⊥ Q M$ . Thus the circle on diameter $O P$ is $x (x - x_{1}) + y (y - y_{1}) = 0$ , and the midpoint of any chord from $P$ lies on the circle with diameter $O P$ scaled by half, or more simply, the midpoint $M (h, - h)$ satisfies the condition that $M$ lies strictly inside the circle and lies on the locus of midpoints of chords through $P$ . Let's use the quadratic equation in the chord's slope or the midpoint coordinate $h$ .
Execute: Note that the circle is $x^{2} + y^{2} - x_{1} x - y_{1} y = 0$ , where $x_{1} = \frac{1 + a}{2}$ and $y_{1} = \frac{1 - a}{2}$ . The center is $C = (\frac{x_{1}}{2}, \frac{y_{1}}{2})$ . Let $M (h, - h)$ be the midpoint of a chord from $P$ . The equation of the chord having midpoint $M (h, k)$ is $T = S_{1}$ : $x h + y k - \frac{x_{1}}{2} (x + h) - \frac{y_{1}}{2} (y + k) = h^{2} + k^{2} - x_{1} h - y_{1} k$ . Since $P (x_{1}, y_{1})$ lies on this chord: $x_{1} h + y_{1} k - \frac{x_{1}}{2} (x_{1} + h) - \frac{y_{1}}{2} (y_{1} + k) = h^{2} + k^{2} - x_{1} h - y_{1} k$ . Simplify this: $\frac{1}{2} x_{1} h + \frac{1}{2} y_{1} k - \frac{1}{2} (x_{1}^{2} + y_{1}^{2}) = h^{2} + k^{2} - x_{1} h - y_{1} k$ $h^{2} + k^{2} - \frac{3}{2} x_{1} h - \frac{3}{2} y_{1} k + \frac{1}{2} (x_{1}^{2} + y_{1}^{2}) = 0$ . Wait, simpler: the midpoint $M$ of chord $P Q$ where $P$ is fixed and $Q$ is on the circle $x^{2} + y^{2} - x_{1} x - y_{1} y = 0$ satisfies $Q = 2 M - P$ . Since $Q$ is on the circle: $(2 h - x_{1})^{2} + (2 k - y_{1})^{2} - x_{1} (2 h - x_{1}) - y_{1} (2 k - y_{1}) = 0$ $4 h^{2} - 4 h x_{1} + x_{1}^{2} + 4 k^{2} - 4 k y_{1} + y_{1}^{2} - 2 h x_{1} + x_{1}^{2} - 2 k y_{1} + y_{1}^{2} = 0$ $4 (h^{2} + k^{2}) - 6 x_{1} h - 6 y_{1} k + 2 (x_{1}^{2} + y_{1}^{2}) = 0$ $2 (h^{2} + k^{2}) - 3 x_{1} h - 3 y_{1} k + (x_{1}^{2} + y_{1}^{2}) = 0$ . Since $M$ lies on $x + y = 0$ , substitute $k = - h$ : $2 (h^{2} + h^{2}) - 3 x_{1} h + 3 y_{1} h + (x_{1}^{2} + y_{1}^{2}) = 0$ $4 h^{2} - 3 (x_{1} - y_{1}) h + (x_{1}^{2} + y_{1}^{2}) = 0$ .
Execute: Substitute $x_{1} = \frac{1 + a}{2}$ and $y_{1} = \frac{1 - a}{2}$ : $x_{1} - y_{1} = \frac{(1 + a) - (1 - a)}{2} = a$ $x_{1}^{2} + y_{1}^{2} = \frac{(1 + a)^{2} + (1 - a)^{2}}{4} = \frac{2 (1 + a^{2})}{4} = \frac{1 + a^{2}}{2}$ . Thus the quadratic equation in $h$ is: $4 h^{2} - 3 a h + \frac{1 + a^{2}}{2} = 0$ , which is $8 h^{2} - 6 a h + (1 + a^{2}) = 0$ . For two distinct chords, there must be two distinct midpoints $M$ (and neither can be $P$ itself, since a chord has two endpoints and its midpoint cannot be an endpoint unless it's degenerate). For two distinct real roots of $h$ , the discriminant must be strictly positive: $D > 0 ⟹ (- 6 a)^{2} - 4 (8) (1 + a^{2}) > 0$ $36 a^{2} - 32 (1 + a^{2}) > 0$ $36 a^{2} - 32 - 32 a^{2} > 0$ $4 a^{2} - 32 > 0 ⟹ 4 a^{2} > 32 ⟹ a^{2} > 8$ .
Verify: Check if $M = P$ can be a root: $M = P ⟹ h = x_{1} = \frac{1 + a}{2}$ and $- h = y_{1} = \frac{1 - a}{2} ⟹ x_{1} + y_{1} = 0 ⟹ 1 = 0$ , impossible! So $P$ never lies on $x + y = 0$ , meaning the chords are never degenerate tangents. Thus, $a^{2} > 8$ completely guarantees two distinct real chords, giving $a^{2} \in (8, \infty)$ .

Condition for Chords from a Point on a Circle to be Bisected by a Line

What feels right?

The line $x + y = 0$ bisects two distinct chords drawn from $P$ if and only if the resulting quadratic equation in the chord midpoint coordinate has a positive discriminant, yielding $a^{2} \in (8, \infty)$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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