JEE MainMathematics

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Circles: JEE Main Mathematics Question with Solution

Q: Why is the other endpoint ?

Because is defined as the midpoint of chord , by the midpoint formula , which rearranges to .

The set of all values of

a^2

for which the line

x + y = 0

bisects two distinct chords drawn from a point

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

on the circle

2x^2 + 2y^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 $\Delta > 0$ to $\Delta \le 0$ or assumed the interval was bounded above. Ensure that the condition for two distinct real roots requires $\Delta > 0$ , 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)

None. 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 $C: 2x^2 + 2y^2 - (1 + a)x - (1 - a)y = 0$ , point $P(x_1, y_1) = \left(\frac{1+a}{2}, \frac{1-a}{2}\right)$ 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 $PQ$ 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 $PQ$ where line $OP$ 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 $PM \perp QM$ . Thus the circle on diameter $OP$ is $x(x-x_1) + y(y-y_1) = 0$ , and the midpoint of any chord from $P$ lies on the circle with diameter $OP$ 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 $C = \left(\frac{x_1}{2}, \frac{y_1}{2}\right)$ . 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 $PQ$ where $P$ is fixed and $Q$ is on the circle $x^2 + y^2 - x_1 x - y_1 y = 0$ satisfies $Q = 2M - P$ . Since $Q$ is on the circle: $(2h - x_1)^2 + (2k - y_1)^2 - x_1(2h - x_1) - y_1(2k - y_1) = 0$ $4h^2 - 4h x_1 + x_1^2 + 4k^2 - 4k y_1 + y_1^2 - 2h x_1 + x_1^2 - 2k y_1 + y_1^2 = 0$ $4(h^2 + k^2) - 6x_1 h - 6y_1 k + 2(x_1^2 + y_1^2) = 0$ $2(h^2 + k^2) - 3x_1 h - 3y_1 k + (x_1^2 + y_1^2) = 0$ . Since $M$ lies on $x + y = 0$ , substitute $k = -h$ : $2(h^2 + h^2) - 3x_1 h + 3y_1 h + (x_1^2 + y_1^2) = 0$ $4h^2 - 3(x_1 - y_1)h + (x_1^2 + y_1^2) = 0$ .

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: $4h^2 - 3ah + \frac{1 + a^2}{2} = 0$ , which is $8h^2 - 6ah + (1 + a^2) = 0$ . 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: $D > 0 \implies (-6a)^2 - 4(8)(1 + a^2) > 0$ $36a^2 - 32(1 + a^2) > 0$ $36a^2 - 32 - 32a^2 > 0$ $4a^2 - 32 > 0 \implies 4a^2 > 32 \implies a^2 > 8$ .

✓verify

Check if $M = P$ can be a root: $M = P \implies h = x_1 = \frac{1+a}{2}$ and $-h = y_1 = \frac{1-a}{2} \implies x_1 + y_1 = 0 \implies 1 = 0$ , 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 $x^2 + y^2 - \frac{1+a}{2}x - \frac{1-a}{2}y = 0$ ?
$hx - hy - \frac{1+a}{4}(x + h) - \frac{1-a}{4}(y - h) = 2h^2 - \frac{1+a}{2}h + \frac{1-a}{2}h$
Since the chord passes through $P\left(\frac{1+a}{2}, \frac{1-a}{2}\right)$ , substituting $P$ into $T = S_1$ yields which quadratic equation in $h$ ?
$8h^2 - 4ah + 2 = 0$ , which simplifies to $4h^2 - 2ah + 1 = 0$
What condition must $4h^2 - 2ah + 1 = 0$ satisfy so that there are two distinct chords?
The discriminant must be strictly positive: $(-2a)^2 - 4(4)(1) > 0$ , giving $a^2 > 8$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the other endpoint

Q = 2M - P

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

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 $\Delta > 0$ to $\Delta \le 0$ or assumed the interval was bounded above. Ensure that the condition for two distinct real roots requires $\Delta > 0$ , 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: $(8, \infty)$ - None. 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 $C: 2x^2 + 2y^2 - (1 + a)x - (1 - a)y = 0$ , point $P(x_1, y_1) = \left(\frac{1+a}{2}, \frac{1-a}{2}\right)$ 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 $(4, \infty)$ lying on $x + y = 0$ be $M(h, -h)$ . Since $M$ is the midpoint of chord $x+y=0$ 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 $PQ$ where line $OP$ 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 $PM \perp QM$ . Thus the circle on diameter $OP$ is $x(x-x_1) + y(y-y_1) = 0$ , and the midpoint of any chord from $P$ lies on the circle with diameter $OP$ scaled by half, or more simply, the midpoint $h$ 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 $C = \left(\frac{x_1}{2}, \frac{y_1}{2}\right)$ . 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 $PQ$ where $P$ is fixed and $Q$ is on the circle $x^2 + y^2 - x_1 x - y_1 y = 0$ satisfies $Q = 2M - P$ . Since $Q$ is on the circle: $(2h - x_1)^2 + (2k - y_1)^2 - x_1(2h - x_1) - y_1(2k - y_1) = 0$ $4h^2 - 4h x_1 + x_1^2 + 4k^2 - 4k y_1 + y_1^2 - 2h x_1 + x_1^2 - 2k y_1 + y_1^2 = 0$ $4(h^2 + k^2) - 6x_1 h - 6y_1 k + 2(x_1^2 + y_1^2) = 0$ $2(h^2 + k^2) - 3x_1 h - 3y_1 k + (x_1^2 + y_1^2) = 0$ . Since $M$ lies on $x + y = 0$ , substitute $k = -h$ : $2(h^2 + h^2) - 3x_1 h + 3y_1 h + (x_1^2 + y_1^2) = 0$ $4h^2 - 3(x_1 - y_1)h + (x_1^2 + y_1^2) = 0$ .
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: $4h^2 - 3ah + \frac{1 + a^2}{2} = 0$ , which is $8h^2 - 6ah + (1 + a^2) = 0$ . 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: $D > 0 \implies (-6a)^2 - 4(8)(1 + a^2) > 0$ $36a^2 - 32(1 + a^2) > 0$ $36a^2 - 32 - 32a^2 > 0$ $4a^2 - 32 > 0 \implies 4a^2 > 32 \implies a^2 > 8$ .
verify: Check if $M = P$ can be a root: $M = P \implies h = x_1 = \frac{1+a}{2}$ and $-h = y_1 = \frac{1-a}{2} \implies x_1 + y_1 = 0 \implies 1 = 0$ , 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 $a^2 \in (8, \infty)$ .

Circles: JEE Main Mathematics Question with Solution

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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