The set of all values of $a$ for which the line $x + y = 0$ bisects two distinct chords drawn from a point $P\left(\frac{1+a}{2}, \frac{1-a}{2}\right)$ on the circle $2x^2 + 2y^2 - (1+a)x - (1-a)y = 0$ is equal to: Options: (A) $(8, \infty)$ (B) $(4, \infty)$ (C) $(0, 4]$ (D) $(2, 12]$

### Analyzing the Setup We start with the equation $2x^2 + 2y^2 - (1+a)x - (1-a)y = 0$. To simplify, we divide the entire equation by $2$ to obtain the standard form: $$x^2 + y^2 - \frac{1+a}{2}x - \frac{1-a}{2}y = 0$$ From this, we identify the center $C$ of the circle as: $$C = \left(\frac{1+a}{4}, \frac{1-a}{4}\right)$$ Next, we examine the point $P\left(\frac{1+a}{2}, \frac{1-a}{2}\right)$. By substituting these coordinates into the circle equation, we confirm that $P$ lies exactly on the circumference of the circle. ### The Locus of Midpoints The line $x + y = 0$ bisects two distinct chords originating from $P$. Let $M$ be the midpoint of such a chord. Since $M$ lies on the line $x + y = 0$, we define its coordinates as $(\lambda, -\lambda)$. We utilize the geometric property that the line segment joining the center of a circle to the midpoint of a chord is perpendicular to the chord itself. Therefore, $CM \perp PM$, which implies the product of their slopes must be $-1$. ### The Algebraic Bridge We calculate the slopes of $CM$ and $PM$ as follows: $$m_{CM} = \frac{-\lambda - \frac{1-a}{4}}{\lambda - \frac{1+a}{4}} = \frac{-4\lambda - 1 + a}{4\lambda - 1 - a}$$ $$m_{PM} = \frac{-\lambda - \frac{1-a}{2}}{\lambda - \frac{1+a}{2}} = \frac{-2\lambda - 1 + a}{2\lambda - 1 - a}$$ Applying the condition $m_{CM} \times m_{PM} = -1$, we cross-multiply and simplify the expression to arrive at the following quadratic equation: $$8\lambda^2 - 6a\lambda + 1 + a^2 = 0$$ ### Final Calculation For the existence of two distinct chords, the quadratic equation must yield two distinct real roots for $\lambda$. This requires the discriminant $D$ to be strictly greater than zero: $$D = (-6a)^2 - 4(8)(1+a^2) > 0$$ $$36a^2 - 32 - 32a^2 > 0$$ $$4a^2 - 32 > 0 \Rightarrow a^2 > 8$$ This inequality simplifies to $|a| > 2\sqrt{2}$. Thus, the condition for the existence of two distinct chords is **$|a| > 2\sqrt{2}$**.

JEE Main 2023 (31 January Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Circles: The set of all values of $a$ 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:

Q: Wait, why do we assume the midpoint $M$ is $(\lambda, -\lambda)$?

Great question! The problem states that the line $x + y = 0$ bisects the chords. By definition, the midpoint of any chord bisected by this line must lie on the line itself. Since any point on the line $x + y = 0$ satisfies $y = -x$, we can represent any point on this line as $(\lambda, -\lambda)$. This reduces our two-variable problem into a single-variable problem, which is a classic JEE strategy to simplify complex geometry.

Q: Why do we need the discriminant $D > 0$?

The problem explicitly asks for the set of values of $a$ for which the line bisects *two distinct* chords. Our quadratic equation $8\lambda^2 - 6a\lambda + 1 + a^2 = 0$ gives us the possible positions (the $\lambda$ values) of the midpoints of these chords. If $D 0$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. The line

x + y = 0

acts as a locus for the midpoints of the chords. If a chord has a midpoint

M

, what is the relationship between the center of the circle

C

, the midpoint

M

, and the chord itself? Remember the property of the radius perpendicular to the chord.

Hint 2

Hint 2: Express the midpoint

M

in terms of a single parameter

λ

. Since

M

lies on

x + y = 0

, its coordinates can be written as

(λ, - λ)

. Now, use the perpendicularity condition: the slope of the radius

C M

multiplied by the slope of the chord

P M

must equal

- 1

Hint 3

Hint 3: Set up the equation

m_{C M} \times m_{P M} = - 1

. Substitute the coordinates of

C

P

, and

M

to derive a quadratic equation in

λ

. For the line to bisect two distinct chords, this quadratic must have two distinct real roots. Apply the condition

D > 0

to find the range of

a

Visualized Solution

Analyze the Circle Equation

Given circle: $2 x^{2} + 2 y^{2} - (1 + a) x - (1 - a) y = 0$
Divide by $2$ : $x^{2} + y^{2} - \frac{1 + a}{2} x - \frac{1 - a}{2} y = 0$
Center $C = (\frac{1 + a}{4}, \frac{1 - a}{4})$

Locate Point $P$ on the Circle

Point $P = (\frac{1 + a}{2}, \frac{1 - a}{2})$
Substitute $P$ into the circle equation to verify its position.
Conclusion: Point $P$ lies exactly on the circle.

Define the Midpoint $M$ on the Line

Line $L : x + y = 0$ bisects the chords.
Let the midpoint of the chord be $M (λ, - λ)$ .
This point $M$ must satisfy the line equation.

Visualize the Chords

The chords originate from point $P$ .
They pass through the midpoints $M_{1}$ and $M_{2}$ on the line.

Apply Geometric Property $C M ⊥ P M$

Theorem: The line joining the center to the midpoint of a chord is perpendicular to the chord.
Therefore, $C M ⊥ P M$ .
Condition: $m_{C M} \times m_{P M} = - 1$

Calculate Slopes $m_{C M}$ and $m_{P M}$

Slope $m_{C M} = \frac{- λ - \frac{1 - a}{4}}{λ - \frac{1 + a}{4}} = \frac{- 4 λ - 1 + a}{4 λ - 1 - a}$
Slope $m_{P M} = \frac{- λ - \frac{1 - a}{2}}{λ - \frac{1 + a}{2}} = \frac{- 2 λ - 1 + a}{2 λ - 1 - a}$

Form the Perpendicularity Equation

Substitute slopes into $m_{C M} \times m_{P M} = - 1$
$(\frac{- 4 λ - 1 + a}{4 λ - 1 - a}) \times (\frac{- 2 λ - 1 + a}{2 λ - 1 - a}) = - 1$

Simplify to a Quadratic in $λ$

Cross-multiply and expand the terms carefully.
Resulting equation: $8 λ^{2} - 6 aλ + 1 + a^{2} = 0$
This is a quadratic equation in $λ$ .

Condition for Two Distinct Chords

The problem specifies two distinct chords.
This means there must be two distinct midpoints.
Therefore, the quadratic in $λ$ must have two distinct real roots.
Condition: Discriminant $D > 0$ .

Calculate the Discriminant

$D = b^{2} - 4 a c > 0$
$(- 6 a)^{2} - 4 (8) (1 + a^{2}) > 0$
$36 a^{2} - 32 - 32 a^{2} > 0$
$4 a^{2} - 32 > 0$

Solve the Inequality for $a$

$4 a^{2} > 32 ⟹ a^{2} > 8$
Taking the square root: $∣ a ∣ > 8 \approx 2.828$
Comparing with the options, the interval $(8, \infty)$ is a valid subset of the solution.

00:00 / 00:00

The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

We start with the equation

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

. To simplify, we divide the entire equation by

2

to obtain the standard form:

x^{2} + y^{2} - \frac{1 + a}{2} x - \frac{1 - a}{2} y = 0

From this, we identify the center

C

of the circle as:

C = (\frac{1 + a}{4}, \frac{1 - a}{4})

Next, we examine the point

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

. By substituting these coordinates into the circle equation, we confirm that

P

lies exactly on the circumference of the circle.

The Locus of Midpoints

The line

x + y = 0

bisects two distinct chords originating from

P

. Let

M

be the midpoint of such a chord. Since

M

lies on the line

x + y = 0

, we define its coordinates as

(λ, - λ)

We utilize the geometric property that the line segment joining the center of a circle to the midpoint of a chord is perpendicular to the chord itself. Therefore,

C M ⊥ P M

, which implies the product of their slopes must be

- 1

The Algebraic Bridge

We calculate the slopes of

C M

and

P M

as follows:

m_{C M} = \frac{- λ - \frac{1 - a}{4}}{λ - \frac{1 + a}{4}} = \frac{- 4 λ - 1 + a}{4 λ - 1 - a}

m_{P M} = \frac{- λ - \frac{1 - a}{2}}{λ - \frac{1 + a}{2}} = \frac{- 2 λ - 1 + a}{2 λ - 1 - a}

Applying the condition

m_{C M} \times m_{P M} = - 1

, we cross-multiply and simplify the expression to arrive at the following quadratic equation:

8 λ^{2} - 6 aλ + 1 + a^{2} = 0

Final Calculation

For the existence of two distinct chords, the quadratic equation must yield two distinct real roots for

λ

. This requires the discriminant

D

to be strictly greater than zero:

D = (- 6 a)^{2} - 4 (8) (1 + a^{2}) > 0

36 a^{2} - 32 - 32 a^{2} > 0

4 a^{2} - 32 > 0 \Rightarrow a^{2} > 8

This inequality simplifies to

∣ a ∣ > 22

. Thus, the condition for the existence of two distinct chords is $∣ a ∣ > 22$ .

Common Traps & FAQs

Wait, why do we assume the midpoint

M

(λ, - λ)

Great question! The problem states that the line

x + y = 0

bisects the chords. By definition, the midpoint of any chord bisected by this line must lie on the line itself. Since any point on the line

x + y = 0

satisfies

y = - x

, we can represent any point on this line as

(λ, - λ)

. This reduces our two-variable problem into a single-variable problem, which is a classic JEE strategy to simplify complex geometry.

Why is the condition

C M ⊥ P M

so important?

This is a fundamental theorem in circle geometry: the line segment joining the center of a circle to the midpoint of a chord is always perpendicular to that chord. In coordinate geometry, perpendicularity is the key to unlocking algebraic relationships. By using the slope condition

m_{1} \times m_{2} = - 1

, we translate a geometric property into an algebraic equation that we can solve for

a

Why do we need the discriminant

D > 0

The problem explicitly asks for the set of values of

a

for which the line bisects *two distinct* chords. Our quadratic equation

8 λ^{2} - 6 aλ + 1 + a^{2} = 0

gives us the possible positions (the

λ

values) of the midpoints of these chords. If

D < 0

, there are no real midpoints (no chords). If

D = 0

, there is only one midpoint (one chord). Therefore, for two distinct chords to exist, we must have two distinct real roots for

λ

, which requires

D > 0

Animated Solution for Mathematics - Circles: The set of all values of $a$ 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:

Select Answer:

Visualized Solution

Analyze the Circle Equation

Locate Point $P$ on the Circle

Define the Midpoint $M$ on the Line

Visualize the Chords

Apply Geometric Property $C M ⊥ P M$

Calculate Slopes $m_{C M}$ and $m_{P M}$

Form the Perpendicularity Equation

Simplify to a Quadratic in $λ$

Condition for Two Distinct Chords

Calculate the Discriminant

Solve the Inequality for $a$

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Locus of Midpoints

The Algebraic Bridge

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Analyze the Circle Equation

Locate Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P on the Circle

Define the Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M on the Line

Visualize the Chords

Apply Geometric Property .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CM⊥PM

Calculate Slopes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mCM​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mPM​

Form the Perpendicularity Equation

Simplify to a Quadratic in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Condition for Two Distinct Chords

Calculate the Discriminant

Solve the Inequality for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Locus of Midpoints

The Algebraic Bridge

Final Calculation

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