Mirror Image of Circle Formula — Complete Guide for JEE

Introduction

The mirror image of a circle with respect to a line is one of the most frequently asked concepts in JEE Main and Advanced Coordinate Geometry. The idea is simple: reflect the circle across a given line (the "mirror"), and write the equation of the new circle. Since reflection preserves distances, the radius stays the same — only the center changes.

This article covers the complete formula, step-by-step method, common mirror lines (axes, $y = x$, arbitrary lines), and solved JEE problems.

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The Core Idea

When a circle is reflected across a line:

1. The radius remains unchanged — reflection is a rigid transformation
2. The center gets reflected to a new position across the line
3. The image circle has the same radius but a new center

So the entire problem reduces to: find the mirror image of the center.

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Mirror Image of a Point — The Foundation Formula

To find the mirror image of a point $(h, k)$ in the line $ax + by + c = 0$, use:

$\boxed{\frac{x - h}{a} = \frac{y - k}{b} = \frac{-2(ah + bk + c)}{a^2 + b^2}}$

This gives the reflected point $(x, y)$ directly.

Derivation (Why It Works)

Two conditions define the mirror image:

Condition 1 — Perpendicularity: The line joining $(h, k)$ and its image $(x, y)$ is perpendicular to the mirror line $ax + by + c = 0$.

The mirror line has slope $-a/b$, so the joining line has slope $b/a$:

$\frac{y - k}{x - h} = \frac{b}{a} \implies \frac{x - h}{a} = \frac{y - k}{b}$

Condition 2 — Midpoint on the line: The midpoint $\left(\frac{x+h}{2}, \frac{y+k}{2}\right)$ lies on $ax + by + c = 0$:

$a\left(\frac{x+h}{2}\right) + b\left(\frac{y+k}{2}\right) + c = 0$

Solving these two conditions together gives the formula.

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Mirror Image of a Circle — Step-by-Step Method

Given:

• Circle: $x^2 + y^2 + 2gx + 2fy + c = 0$ with center $(-g, -f)$ and radius $r = \sqrt{g^2 + f^2 - c}$
• Mirror line: $ax + by + d = 0$

Step 1: Find the center of the original circle

$\text{Center} = (-g, -f), \quad r = \sqrt{g^2 + f^2 - c}$

Step 2: Reflect the center across the mirror line

Using the reflection formula with point $(-g, -f)$ and line $ax + by + d = 0$:

$\frac{x - (-g)}{a} = \frac{y - (-f)}{b} = \frac{-2(a(-g) + b(-f) + d)}{a^2 + b^2}$

Let $\lambda = \frac{-2(-ag - bf + d)}{a^2 + b^2} = \frac{2(ag + bf - d)}{a^2 + b^2}$

Then the new center is: $\text{New Center} = (-g + a\lambda, \; -f + b\lambda)$

Step 3: Write the equation of the image circle

$\boxed{(x - x_1)^2 + (y - y_1)^2 = r^2}$

where $(x_1, y_1)$ is the reflected center and $r$ is the same radius as the original.

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Quick Formulas for Common Mirror Lines

Mirror Line: x-axis ($y = 0$)

Reflect center $(h, k)$: image is $(h, -k)$

$\text{If circle is } (x-h)^2 + (y-k)^2 = r^2$ $\text{Image: } (x-h)^2 + (y+k)^2 = r^2$

Rule: Replace $y$ with $-y$ in the center, or equivalently replace $f$ with $-f$ in general form.

Mirror Line: y-axis ($x = 0$)

Reflect center $(h, k)$: image is $(-h, k)$

$\text{Image: } (x+h)^2 + (y-k)^2 = r^2$

Rule: Replace $x$ with $-x$ in the center, or replace $g$ with $-g$.

Mirror Line: $y = x$

Reflect center $(h, k)$: image is $(k, h)$ — swap coordinates.

$\text{Image: } (x-k)^2 + (y-h)^2 = r^2$

Mirror Line: $y = -x$

Reflect center $(h, k)$: image is $(-k, -h)$.

$\text{Image: } (x+k)^2 + (y+h)^2 = r^2$

Mirror Line: $x = a$ (vertical line)

Reflect center $(h, k)$: image is $(2a - h, \; k)$.

Mirror Line: $y = b$ (horizontal line)

Reflect center $(h, k)$: image is $(h, \; 2b - k)$.

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Summary Table

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Solved Examples

Example 1: Mirror image in the x-axis

Q: Find the image of the circle $x^2 + y^2 - 4x + 6y + 4 = 0$ in the $x$-axis.

Solution:

Center = $(2, -3)$, radius $= \sqrt{4 + 9 - 4} = 3$

Mirror image of center $(2, -3)$ in $x$-axis: $(2, 3)$

Image circle: $(x-2)^2 + (y-3)^2 = 9$

Expanding: $x^2 + y^2 - 4x - 6y + 4 = 0$

Notice: Only the sign of the $y$-term changed ($+6y \to -6y$).

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Example 2: Mirror image in $y = x$

Q: Find the mirror image of $x^2 + y^2 - 2x = 0$ in the line $y = 3 - x$.

Solution:

Circle: center $(1, 0)$, radius $= 1$

Mirror line: $x + y - 3 = 0$ (rewrite $y = 3 - x$)

Reflect center $(1, 0)$ in $x + y - 3 = 0$:

$\frac{x - 1}{1} = \frac{y - 0}{1} = \frac{-2(1 + 0 - 3)}{1 + 1} = \frac{-2(-2)}{2} = 2$

So: $x = 1 + 2 = 3$, $y = 0 + 2 = 2$

New center: $(3, 2)$, radius = $1$

Image circle: $(x-3)^2 + (y-2)^2 = 1$

Expanding: $x^2 + y^2 - 6x - 4y + 12 = 0$

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Example 3: General line mirror (JEE-level)

Q: The image of $x^2 + y^2 + 16x - 24y + 183 = 0$ in the line $4x + 7y + 13 = 0$ is?

Solution:

Center = $(-8, 12)$, radius $= \sqrt{64 + 144 - 183} = \sqrt{25} = 5$

Reflect $(-8, 12)$ in $4x + 7y + 13 = 0$:

$\frac{x + 8}{4} = \frac{y - 12}{7} = \frac{-2(4(-8) + 7(12) + 13)}{16 + 49}$

$= \frac{-2(-32 + 84 + 13)}{65} = \frac{-2(65)}{65} = -2$

So: $x = -8 + 4(-2) = -16$, $y = 12 + 7(-2) = -2$

New center: $(-16, -2)$, radius = $5$

Image circle: $(x+16)^2 + (y+2)^2 = 25$

Expanding: $x^2 + y^2 + 32x + 4y + 231 = 0$

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Example 4: Mirror image about origin

Q: Find the image of circle $(x-3)^2 + (y-4)^2 = 25$ about the origin.

Solution:

Reflection about origin means the image of $(h, k)$ is $(-h, -k)$.

Center $(3, 4) \to (-3, -4)$, radius = $5$

Image: $(x+3)^2 + (y+4)^2 = 25$

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Common Mistakes to Avoid

1. Changing the radius: The radius NEVER changes in reflection. Only the center moves.

2. Using the wrong formula direction: In $\frac{x-h}{a} = \frac{y-k}{b} = \lambda$, the final answer is $x = h + a\lambda$, not $x = a + h\lambda$.

3. Sign errors in the numerator: Double-check $ah + bk + c$ — substitute carefully with signs.

4. Confusing mirror image with foot of perpendicular: The foot of perpendicular is the midpoint between the point and its image. For mirror image, use $-2$ in the formula; for foot, use $-1$.

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Quick Trick: Foot of Perpendicular vs Mirror Image

$\text{Foot of perpendicular: } \frac{x - h}{a} = \frac{y - k}{b} = \frac{-(ah + bk + c)}{a^2 + b^2}$

$\text{Mirror image: } \frac{x - h}{a} = \frac{y - k}{b} = \frac{-2(ah + bk + c)}{a^2 + b^2}$

The only difference is the factor of 2 in the numerator. The mirror image is exactly twice as far from the original point as the foot.

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JEE Pattern Questions

Pattern 1: Self-image circles

If a circle is its own mirror image in a line, the line must pass through the center. This means $ah + bk + c = 0$ where $(h,k)$ is the center and $ax + by + c = 0$ is the mirror line.

Pattern 2: Finding the mirror line

If the original and image circles are given, the mirror line is the perpendicular bisector of the line joining their centers.

Pattern 3: Two reflections

Reflecting a circle first in line $L_1$ and then in line $L_2$ is equivalent to a rotation by $2\theta$ about their intersection point, where $\theta$ is the angle between $L_1$ and $L_2$.

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Practice Problems

Problem 1: Find the image of $x^2 + y^2 - 6x + 8y - 11 = 0$ in the line $y = x$.

Answer: Center $(3, -4) \to (-4, 3)$. Image: $(x+4)^2 + (y-3)^2 = 36$, i.e., $x^2 + y^2 + 8x - 6y - 11 = 0$.

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Problem 2: The image of circle $x^2 + y^2 = 4$ in the line $x + y = 2$ is?

Answer: Center $(0,0)$ reflected in $x + y - 2 = 0$ gives $(2, 2)$. Image: $(x-2)^2 + (y-2)^2 = 4$.

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Problem 3: If the image of $(x-1)^2 + (y-2)^2 = 5$ in a line $L$ is $(x-3)^2 + (y-4)^2 = 5$, find $L$.

Answer: $L$ is the perpendicular bisector of centers $(1,2)$ and $(3,4)$. Midpoint = $(2, 3)$, slope of join = $1$, so $L: y - 3 = -1(x - 2)$, i.e., $x + y = 5$.

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Key Takeaways

1. Mirror image of a circle = same radius + reflected center
2. Master the point reflection formula: $\frac{x-h}{a} = \frac{y-k}{b} = \frac{-2(ah+bk+c)}{a^2+b^2}$
3. For standard lines ($x$-axis, $y$-axis, $y=x$), use the shortcut rules
4. The foot of perpendicular formula differs only by a factor of $2$
5. Always verify: the image center should be equidistant from the mirror line as the original