Let a circle C₁ be obtained on rolling the circle x² + y² - 4x - 6y + 11 = 0 upwards 4 units on the

Q: Why do we only reflect the center to find $C_2$ instead of the whole circle?

That is a brilliant question! Remember that a circle is defined entirely by its center and its radius. When you reflect a circle across a line, the radius remains invariant—it does not change size. Therefore, the reflected circle $C_2$ has the same radius as $C_1$. By finding the reflection of the center, you have effectively located the entire circle in the coordinate plane.

Q: How do I know the tangent slope is $1$?

Look at the equation of the tangent we derived: $x - y - 1 = 0$. If you rearrange this into the slope-intercept form $y = mx + c$, you get $y = x - 1$. Here, the coefficient of $x$ is $1$, so the slope $m = 1$. Since $m = \tan \theta$, we have $\tan \theta = 1$, which tells us the angle of inclination is $45^\circ$.

Q: Is the area of the trapezium simply $\frac{1}{2}(y_A + y_B) \times |x_A - x_B|$?

Exactly right! Since the feet of the perpendiculars $M$ and $N$ lie on the $x$-axis, the segments $AM$ and $BN$ are vertical. These are your parallel sides. The distance between them is the horizontal separation of the centers, which is $|x_A - x_B|$. You have perfectly grasped the geometry of the situation.

Question

Let a circle $C_1$ be obtained on rolling the circle $x^2 + y^2 - 4x - 6y + 11 = 0$ upwards 4 units on the tangent $T$ to it at the point $(3, 2)$. Let $C_2$ be the image of $C_1$ in $T$. Let $A$ and $B$ be the centers of circles $C_1$ and $C_2$ respectively, and $M$ and $N$ be respectively the feet of perpendiculars drawn from $A$ and $B$ on the $x$-axis. Then the area of the trapezium $AMNB$ is :

Options:

(A) $2(2 + \sqrt{2})$

(B) $4(1 + \sqrt{2})$

(C) $3 + 2\sqrt{2}$

(D) $2(1 + \sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Every great journey begins with understanding where we stand. We are given the circle $C_0$ defined by the equation:

$$x^2 + y^2 - 4x - 6y + 11 = 0$$

To understand its soul, we must complete the square. By grouping the $x$ and $y$ terms, we transform this into:

$$(x-2)^2 + (y-3)^2 = 2$$

Now, the geometry reveals itself: our circle is centered at $O(2, 3)$ with a radius $r = \sqrt{2}$. This is our starting point.

### The Tangent as a Stage

Next, we encounter the tangent $T$ at the point $P(3, 2)$. In the language of coordinate geometry, we use the $T=0$ method to find the equation of the tangent at a point on the circle.

Substituting the coordinates into the circle equation, we derive:

$$x(3) + y(2) - 2(x+3) - 3(y+2) + 11 = 0$$

Simplifying this, we arrive at the elegant line:

$$x - y - 1 = 0$$

This line is our stage. Its slope is $1$, which tells us it makes an angle of $45^{\circ}$ with the positive $x$-axis. This angle is the key to our next movement.

### The Roll of the Circle

Imagine the circle rolling $4$ units upwards along this tangent. As the circle rolls, its center $O$ must also move $4$ units parallel to the tangent. This is a vector displacement.

Since the tangent is inclined at $45^{\circ}$, the displacement vector is $(4 \cos 45^{\circ}, 4 \sin 45^{\circ})$. Calculating this, we get $(2\sqrt{2}, 2\sqrt{2})$.

Adding this to our original center $O(2, 3)$, we find the new center $A$ of circle $C_1$:

$$A = (2 + 2\sqrt{2}, 3 + 2\sqrt{2})$$

### The Mirror Reflection

Now, the problem introduces a twist: circle $C_2$ is the image of $C_1$ in the tangent $T$. The tangent acts as a mirror. To find the center $B$ of $C_2$, we must reflect point $A$ across the line $x - y - 1 = 0$.

We use the reflection formula:

$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$$

Substituting our coordinates and the line equation, the math simplifies beautifully. The right-hand side evaluates to $2$. Solving for $x$ and $y$, we find the center $B$ to be:

$$B = (4 + 2\sqrt{2}, 1 + 2\sqrt{2})$$

### The Final Trapezium

We are almost there. We drop perpendiculars from $A$ and $B$ onto the $x$-axis to get points $M$ and $N$. These are simply the projections of the centers onto the $x$-axis.

Thus, $M = (2 + 2\sqrt{2}, 0)$ and $N = (4 + 2\sqrt{2}, 0)$. The trapezium $AMNB$ has parallel sides of lengths $h_1 = 3 + 2\sqrt{2}$ and $h_2 = 1 + 2\sqrt{2}$, and a width (the distance between the parallel sides) of:

$$MN = (4 + 2\sqrt{2}) - (2 + 2\sqrt{2}) = 2$$

Using the area formula for a trapezium, $	ext{Area} = \frac{1}{2} 	imes (h_1 + h_2) 	imes 	ext{width}$, we calculate:

$$	ext{Area} = \frac{1}{2} 	imes ((3 + 2\sqrt{2}) + (1 + 2\sqrt{2})) 	imes 2$$

The $2$ and the $\frac{1}{2}$ cancel out, leaving us with $4 + 4\sqrt{2}$, or:

$$\mathbf{4(1 + \sqrt{2})}$$

And there it is. Through careful steps, we have navigated the geometry and arrived at the solution. Remember, in JEE Advanced, it is not just about the final number; it is about the elegance of the path you take to get there.

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Visualized Solution

Equation of Circle $C_{0}$

Equation of Tangent $T$

Slope and Inclination of $T$

Displacement of the Center

Coordinates of Center $A$

Finding Center $B$ of Circle $C_{2}$

Substituting Values for Image

Evaluating Center $B$

Coordinates of Feet of Perpendiculars

Dimensions of Trapezium $A M N B$

Area Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent as a Stage

The Roll of the Circle

The Mirror Reflection

The Final Trapezium

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Visualized Solution

Equation of Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C0​

Equation of Tangent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Slope and Inclination of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T