Wait, why do we reject the case where $\alpha = 2$ and $\beta = -1$?

Great observation! In coordinate geometry, constraints like $a \neq b$ are often 'logical filters.' If we assumed $\alpha = 2$, we would find that $a = 0$. If we then assumed $\beta = -1$, we would find that $a = b$. Since the problem explicitly forbids $a = b$, these values create a mathematical contradiction. Always trust the constraints—they are there to guide you to the unique solution.

JEE Main 2023 (25 January Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Circles: The points of intersection of the line $a x + b y = 0, (a \neq = b)$ and the circle $x^{2} + y^{2} - 2 x = 0$ are $A (α, 0)$ and $B (1, β)$ . The image of the circle with $A B$ as a diameter in the line $x + y + 2 = 0$ is :

Q: Is there a faster way to find the center of the circle with diameter $AB$?

Yes! The center of a circle with diameter $AB$ is simply the midpoint of the segment $AB$. If $A = (x_1, y_1)$ and $B = (x_2, y_2)$, the center is $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$. It is much faster than expanding the full diametric equation if you only need the center for a reflection.

Select Answer:

Progressive Hints

Hint 1

The Spark: The constraint

a \neq = b

is your most powerful tool. When you find the intersection points, you will have two potential candidates for

A

and

B

. Use the line equation

a x + b y = 0

to test these candidates; the condition

a \neq = b

will force you to reject the pair that would make

a = b

Hint 2

The Arsenal: Remember the diametric form of a circle:

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0

. Once you have identified the coordinates of

A

and

B

, this formula will give you the equation of the circle instantly. For the reflection, recall that reflecting a circle across a line only changes the center; the radius remains invariant.

Hint 3

The Execution: Start by setting

y = 0

in the circle equation

x^{2} + y^{2} - 2 x = 0

to find the x-coordinates of the intersection points. Once you have

A (0, 0)

and

B (1, 1)

, find the center of the circle with

A B

as the diameter by taking the midpoint of

A B

. Then, apply the reflection formula

\frac{x ^{'} - x _{1}}{a} = \frac{y ^{'} - y _{1}}{b} = - 2 \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}

to the center

C (\frac{1}{2}, \frac{1}{2})

using the line

x + y + 2 = 0

Visualized Solution

Point $A$ on the Circle

Point $A (α, 0)$ lies on the circle $x^{2} + y^{2} - 2 x = 0$ .
Substitute $y = 0$ and $x = α$ .
$α^{2} + 0^{2} - 2 α = 0 ⟹ α (α - 2) = 0$ .
Possible values: $α = 0$ or $α = 2$ .

Eliminating $α = 2$

If $α = 2$ , $A (2, 0)$ lies on $a x + b y = 0 ⟹ 2 a = 0 ⟹ a = 0$ .
Since $a \neq = b$ , $b \neq = 0$ .
Point $B (1, β)$ on $a x + b y = 0 ⟹ 0 (1) + b β = 0 ⟹ β = 0$ .
Check $B (1, 0)$ on $x^{2} + y^{2} - 2 x = 0$ : $1^{2} + 0^{2} - 2 (1) = - 1 \neq = 0$ (Contradiction).
Therefore, $α = 0$ and $A = (0, 0)$ .

Finding Point $B$

Point $B (1, β)$ lies on the circle $x^{2} + y^{2} - 2 x = 0$ .
Substitute $x = 1, y = β$ : $1^{2} + β^{2} - 2 (1) = 0$ .
$β^{2} = 1 ⟹ β = \pm 1$ .

Eliminating $β = - 1$

If $β = - 1$ , the line $a x + b y = 0$ passes through $A (0, 0)$ and $B (1, - 1)$ .
This implies $a (1) + b (- 1) = 0 ⟹ a = b$ .
But we are given $a \neq = b$ . This is not allowed.
So, $β = 1$ and $B = (1, 1)$ .

Circle with Diameter $A B$

We need the equation of the circle with $A B$ as the diameter.
Using diametric form: $(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$ .
Substitute $A (0, 0)$ and $B (1, 1)$ : $(x - 0) (x - 1) + (y - 0) (y - 1) = 0$ .
$x^{2} - x + y^{2} - y = 0 ⟹ x^{2} + y^{2} - x - y = 0$ .

Center and Radius of Diameter Circle

For the circle $x^{2} + y^{2} - x - y = 0$ .
Center $C = (\frac{1}{2}, \frac{1}{2})$ .
Radius $R = (\frac{1}{2})^{2} + (\frac{1}{2})^{2} - 0 = \frac{1}{2}$ .

Reflection of the Center

The circle is reflected across the line $x + y + 2 = 0$ .
The radius remains the same, only the center is reflected.
Let the image of $C (\frac{1}{2}, \frac{1}{2})$ be $C^{'} (x^{'}, y^{'})$ .
Formula: $\frac{x ^{'} - x _{1}}{a} = \frac{y ^{'} - y _{1}}{b} = - 2 \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}$ .

Calculating Image Center $C^{'}$

$\frac{x ^{'} - \frac{1}{2}}{1} = \frac{y ^{'} - \frac{1}{2}}{1} = - 2 \frac{( \frac{1}{2} + \frac{1}{2} + 2 )}{1 ^{2} + 1 ^{2}}$ .
$\frac{x ^{'} - \frac{1}{2}}{1} = - 2 \frac{3}{2} = - 3$ .
$x^{'} = - 3 + \frac{1}{2} = - \frac{5}{2}$ and $y^{'} = - 3 + \frac{1}{2} = - \frac{5}{2}$ .
Image Center $C^{'} = (- \frac{5}{2}, - \frac{5}{2})$ .

Equation of the Image Circle

Image Circle has center $C^{'} (- \frac{5}{2}, - \frac{5}{2})$ and radius $R = \frac{1}{2}$ .
$(x + \frac{5}{2})^{2} + (y + \frac{5}{2})^{2} = (\frac{1}{2})^{2}$ .
$x^{2} + 5 x + \frac{25}{4} + y^{2} + 5 y + \frac{25}{4} = \frac{1}{2}$ .
$x^{2} + y^{2} + 5 x + 5 y + \frac{50}{4} - \frac{2}{4} = 0 ⟹ x^{2} + y^{2} + 5 x + 5 y + 12 = 0$ .

00:00 / 00:00

The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

The given circle is defined by the equation:

x^{2} + y^{2} - 2 x = 0

We are looking for the intersection points

A

and

B

with the line

a x + b y = 0

. Given that

A

has coordinates

(α, 0)

, we substitute this into the circle equation:

α^{2} + 0^{2} - 2 α = 0 \Rightarrow α (α - 2) = 0

This yields two possibilities:

α = 0

α = 2

The Intersection Mystery

The problem imposes the constraint $a eq b$. If we test

α = 2

, the line

a x + b y = 0

forces

a = 0

If we then search for point

B (1, β)

, we find

β = 0

. Testing

B (1, 0)

in the circle equation gives $1^2 + 0^2 - 2(1) = -1 eq 0$, which is a contradiction.

Therefore,

α

must be

0

, identifying point

A

as the origin

(0, 0)

. By logical deduction, we find point

B

to be

(1, 1)

The Diametric Form

We construct the circle using

A B

as the diameter. The diametric form of a circle is given by:

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0

Substituting

A (0, 0)

and

B (1, 1)

, we obtain:

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

The center

C

of this circle is the midpoint of

A B

, which is

(\frac{1}{2}, \frac{1}{2})

. The radius

R

is the distance from the center to the origin:

R = (\frac{1}{2})^{2} + (\frac{1}{2})^{2} = \frac{1}{2}

The Reflection

We reflect the circle across the line

x + y + 2 = 0

. The radius

R = \frac{1}{2}

remains invariant under reflection. We must find the image of the center

C (\frac{1}{2}, \frac{1}{2})

C^{'} (x^{'}, y^{'})

Using the reflection formula:

\frac{x ^{'} - x _{1}}{a} = \frac{y ^{'} - y _{1}}{b} = - 2 \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}

Substituting

x_{1} = \frac{1}{2}

y_{1} = \frac{1}{2}

a = 1

b = 1

, and

c = 2

\frac{x ^{'} - \frac{1}{2}}{1} = \frac{y ^{'} - \frac{1}{2}}{1} = - 2 \frac{( \frac{1}{2} + \frac{1}{2} + 2 )}{1 ^{2} + 1 ^{2}} = - 3

Solving for

x^{'}

and

y^{'}

, we find the new center

C^{'} = (- \frac{5}{2}, - \frac{5}{2})

Final Calculation

The equation of the reflected circle is:

(x + \frac{5}{2})^{2} + (y + \frac{5}{2})^{2} = (\frac{1}{2})^{2}

Expanding this expression, we arrive at the final result: $x^{2} + y^{2} + 5 x + 5 y + 12 = 0$

Common Traps & FAQs

Wait, why do we reject the case where

α = 2

and

β = - 1

Great observation! In coordinate geometry, constraints like

a \neq = b

are often 'logical filters.' If we assumed

α = 2

, we would find that

a = 0

. If we then assumed

β = - 1

, we would find that

a = b

. Since the problem explicitly forbids

a = b

, these values create a mathematical contradiction. Always trust the constraints—they are there to guide you to the unique solution.

Do I need to find the new radius after reflecting the circle?

Not at all! Reflection is an isometry, meaning it preserves distances. When you reflect a circle across a line, the circle's size and shape remain identical. Only the position of the center changes. You can calculate the radius of the original circle and use that same value for the image circle.

Is there a faster way to find the center of the circle with diameter

A B

Yes! The center of a circle with diameter

A B

is simply the midpoint of the segment

A B

. If

A = (x_{1}, y_{1})

and

B = (x_{2}, y_{2})

, the center is

(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

. It is much faster than expanding the full diametric equation if you only need the center for a reflection.

Animated Solution for Mathematics - Circles: The points of intersection of the line $a x + b y = 0, (a \neq = b)$ and the circle $x^{2} + y^{2} - 2 x = 0$ are $A (α, 0)$ and $B (1, β)$ . The image of the circle with $A B$ as a diameter in the line $x + y + 2 = 0$ is :

Select Answer:

Visualized Solution

Point $A$ on the Circle

Eliminating $α = 2$

Finding Point $B$

Eliminating $β = - 1$

Circle with Diameter $A B$

Center and Radius of Diameter Circle

Reflection of the Center

Calculating Image Center $C^{'}$

Equation of the Image Circle

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Intersection Mystery

The Diametric Form

The Reflection

Final Calculation

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Select Answer:

Visualized Solution

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

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=2

Finding Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β=−1

Circle with Diameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Center and Radius of Diameter Circle

Reflection of the Center

Calculating Image Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C′

Equation of the Image Circle

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Intersection Mystery

The Diametric Form

The Reflection

Final Calculation

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