A rectangle R with end points of the one of its sides as (1, 2) and (3, 6) is inscribed in a circle.

Q: Wait, why do we reject the idea of finding the circle's center $(h, k)$ first? Isn't that the standard way to solve circle problems?

That is a fantastic question! While finding the center $(h, k)$ is a valid path, it is often a 'trap' in competitive exams. It leads to a system of equations that is algebraically heavy and prone to calculation errors. In JEE Advanced, we look for geometric elegance. By recognizing the parallelism between the chord and the diameter, we bypass the need for the center's coordinates entirely. Always look for the 'geometric shortcut' before diving into brute-force algebra.

Q: How can we be so sure that the distance from the center to the side is exactly half of the side BC?

Think about the symmetry of a rectangle inscribed in a circle. The center of the circle is the intersection of the diagonals of the rectangle. If you draw a line from the center perpendicular to side AB, it must bisect side AB and also bisect the opposite side. Because the diameter is parallel to the side, this perpendicular line is essentially the 'width' of the rectangle. The distance from the center to the side is exactly half the length of the adjacent side. It is a beautiful consequence of the rectangle's rotational symmetry!

Q: I got the slope of the diameter as 2, but I'm confused—does the diameter have to pass through the center?

Yes, absolutely! By definition, a diameter is any chord that passes through the center of the circle. The equation $2x - y + 4 = 0$ represents the line containing the diameter. Since the center lies on this line, and the line is parallel to our side AB, the distance from the center to the side AB is simply the perpendicular distance between these two parallel lines. You are thinking exactly in the right direction—keep that geometric intuition sharp!

Question

A rectangle $R$ with end points of the one of its sides as $(1, 2)$ and $(3, 6)$ is inscribed in a circle. If the equation of a diameter of the circle is $2x - y + 4 = 0$, then the area of $R$ is _______.

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: Unlocking the Inscribed Rectangle

Welcome, fellow traveler on this journey through coordinate geometry. Today, we are not just solving a problem; we are uncovering a hidden symmetry.

When you look at a rectangle inscribed in a circle, it is easy to get lost in the variables—the radius, the center, the coordinates of the vertices. But the true master of JEE Advanced knows that the most complex problems often yield to the simplest geometric insights.

Let us break this down, step by step, and see the beauty in the math.

### Phase 1: The Spark of Orientation

We begin with two points, $A(1, 2)$ and $B(3, 6)$. These are the endpoints of one side of our rectangle.

Before we do anything else, let us find the slope of this side. The slope $m_1$ is the change in $y$ over the change in $x$:

$$m_1 = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2$$

This tells us the 'steepness' of our side $AB$. Now, let us find the equation of the line containing this side.

Using the point-slope form, $y - y_1 = m(x - x_1)$, we get $y - 2 = 2(x - 1)$, which simplifies beautifully to $2x - y = 0$. This is the foundation of our rectangle.

### Phase 2: The Geometric Revelation

Now, look at the information provided about the diameter: $2x - y + 4 = 0$. Let us calculate its slope, $m_2$.

Using the standard form $ax + by + c = 0$, the slope is $-\frac{a}{b}$. Thus, $m_2 = -\frac{2}{-1} = 2$.

Stop for a moment. Do you see it? The slope of our side $AB$ is $2$, and the slope of the diameter is also $2$. They are parallel!

This is the 'Aha!' moment. In the world of geometry, parallelism is a gift. It tells us that the diameter is not just some random line; it is perfectly aligned with our rectangle.

Because the diameter passes through the center of the circle, and it is parallel to the side $AB$, the center of the circle must lie exactly halfway between the line $2x - y = 0$ and the line containing the opposite side of the rectangle.

### Phase 3: The Distance Calculation

We need to find the distance between these two parallel lines. This distance, $d$, is the perpendicular gap between the side $AB$ and the diameter.

The formula for the distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is:

$$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$$

Substituting our values, where $c_1 = 0$ and $c_2 = 4$, we get:

$$d = \frac{|4 - 0|}{\sqrt{2^2 + (-1)^2}} = \frac{4}{\sqrt{5}}$$

This distance $d$ is the perpendicular distance from the center of the circle to the side $AB$.

### Phase 4: The Anatomy of the Rectangle

Here is where the visualization becomes crucial. The center of the circle is the midpoint of the rectangle.

If you draw a line from the center perpendicular to the side $AB$, that line segment is exactly half the length of the adjacent side, $BC$. Therefore, the total length of side $BC$ must be $2d$:

$$BC = 2 	imes \frac{4}{\sqrt{5}} = \frac{8}{\sqrt{5}}$$

We are almost there. We have the width of the rectangle. Now, we just need the length of the base, $AB$.

Using the distance formula between $A(1, 2)$ and $B(3, 6)$:

$$AB = \sqrt{(3 - 1)^2 + (6 - 2)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$$

### Phase 5: The Final Synthesis

We have the length $AB = 2\sqrt{5}$ and the width $BC = \frac{8}{\sqrt{5}}$. The area of a rectangle is simply the product of its sides:

$$	ext{Area} = AB 	imes BC = (2\sqrt{5}) 	imes \left(\frac{8}{\sqrt{5}}ight)$$

Watch closely as the $\sqrt{5}$ terms cancel out. It is as if the universe intended for this to be elegant. We are left with $2 	imes 8 = 16$.

And there it is. The **Area is 16**.

I hope you felt the thrill of that cancellation. This problem was not about memorizing formulas; it was about seeing the symmetry, trusting the geometry, and letting the algebra follow the path you laid out. Keep practicing, keep visualizing, and remember: in JEE Advanced, the most elegant solution is usually the right one.

Animated Solution for Mathematics - Circles: A rectangle $R$ with end points of the one of its sides as $(1, 2)$ and $(3, 6)$ is inscribed in a circle. If the equation of a diameter of the circle is $2 x - y + 4 = 0$ , then the area of $R$ is _______.

Enter Numerical Value:

Visualized Solution

Visualizing the Given Points

Finding the Slope of Side $A B$

Equation of Line $A B$

Analyzing the Diameter

Parallel Relationship

Distance Between Parallel Lines

Calculating Distance $d$

Relating Distance to Rectangle Side

Calculating Length of Side $B C$

Calculating Length of Side $A B$

Evaluating Length $A B$

Final Area Calculation

Concluding the Area

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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