Let C be the centre of the circle x² + y² - x + 2y = 11/4 and P be a point on the circle. A line pas

Q: Wait, how do we know for sure that $QR$ is a diameter?

That is a fantastic question! In geometry, we must be precise. The problem states that the line passes through the center $C$ and intersects the circle at points $Q$ and $R$. By definition, any chord that passes through the center of a circle is a diameter. It is a simple definition, but in the heat of a JEE Advanced exam, it is easy to overlook these fundamental properties. Always look for the center!

Q: Why can't we just use the coordinate geometry formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$?

You absolutely could! But think about the time cost and the risk. To use that formula, you would first need to find the coordinates of $Q$ and $R$ by solving the intersection of a line and a circle. That involves solving a quadratic equation, which is prone to calculation errors. The geometric approach using $\text{Area} = \frac{1}{2}ab \sin(\theta)$ is elegant, faster, and much less likely to lead to a silly mistake. In JEE, efficiency is just as important as accuracy.

Q: I am confused about $\sin(\frac{3\pi}{4})$. Is it positive or negative?

Great catch! Remember the ASTC rule (All-Sine-Tan-Cos). The angle $\frac{3\pi}{4}$ lies in the second quadrant, where Sine is positive. So, $\sin(\frac{3\pi}{4}) = \sin(\pi - \frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. Keep that unit circle visualization sharp, and you will never get stuck on signs again!

Question

Let $C$ be the centre of the circle $x^2 + y^2 - x + 2y = \frac{11}{4}$ and $P$ be a point on the circle. A line passes through the point $C$, makes an angle of $\frac{\pi}{4}$ with the line $CP$ and intersects the circle at the points $Q$ and $R$. Then the area of the triangle $PQR$ (in unit$^2$) is :

Options:

(A) 2

(B) $2\sqrt{2}$

(C) $8\sin(\frac{\pi}{8})$

(D) $8\cos(\frac{\pi}{8})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a geometric puzzle. Many students see an equation like $x^2 + y^2 - x + 2y = \frac{11}{4}$ and immediately reach for their calculators or start panicking about coordinate geometry.

But I want you to pause. Take a breath. Geometry is not about brute force; it is about visualization and seeing the hidden structure beneath the numbers.

### The Anatomy of the Circle

Our journey begins by decoding the circle. The equation provided is in a messy, expanded form. To understand the circle, we need its heart—the center $C$—and its reach—the radius $r$. We achieve this by completing the square.

We group the $x$ terms $(x^2 - x)$ and the $y$ terms $(y^2 + 2y)$. To complete the square for $x$, we take half of the coefficient of $x$ (which is $-1$), square it to get $\frac{1}{4}$, and add it to both sides. For $y$, we take half of $2$, square it to get $1$, and add that as well.

Our equation transforms into:

$$(x - \frac{1}{2})^2 + (y + 1)^2 = \frac{11}{4} + \frac{1}{4} + 1$$

Simplifying the right-hand side, we get $4$, which is $2^2$. Now, the circle reveals itself: its center $C$ is at $(\frac{1}{2}, -1)$ and its radius $r$ is $2$. This is the foundation of our entire problem.

### The Geometric Setup

Now, imagine the circle on your coordinate plane. We have a point $P$ on the circumference. We have a line passing through the center $C$ that hits the circle at $Q$ and $R$.

Because this line passes through the center, $QR$ is not just any chord; it is a diameter. This is a crucial realization. We are given that the angle between the line $CP$ and the diameter $QR$ is $\frac{\pi}{4}$.

This means $\angle PCQ = \frac{\pi}{4}$. Since $QCR$ is a straight line (a diameter), the angle $\angle PCR$ must be the supplement of $\angle PCQ$. Therefore, $\angle PCR = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

### The Triangle Decomposition

We need the area of $	riangle PQR$. Looking at the diagram, you might be tempted to use base and height, but that would require finding the coordinates of $P, Q,$ and $R$. That is a trap!

Instead, look at the center $C$. Notice how the line segment $CP$ splits the large triangle $	riangle PQR$ into two smaller, more manageable triangles: $	riangle PCQ$ and $	riangle PCR$.

The area of the total triangle is simply the sum of these two parts:

$$	ext{Area}(	riangle PQR) = 	ext{Area}(	riangle PCQ) + 	ext{Area}(	riangle PCR)$$

### The Final Calculation

For any triangle, if we know two sides and the included angle, the area is given by $\frac{1}{2}ab \sin(	heta)$. In both triangles, the sides are radii of the circle. Thus, $CP = CQ = CR = r = 2$.

For $	riangle PCQ$:

$$	ext{Area}(	riangle PCQ) = \frac{1}{2} \cdot CP \cdot CQ \cdot \sin(\frac{\pi}{4}) = \frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2}$$

For $	riangle PCR$:

$$	ext{Area}(	riangle PCR) = \frac{1}{2} \cdot CP \cdot CR \cdot \sin(\frac{3\pi}{4}) = \frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2}$$

Adding these together, we get $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$.

### Conclusion

Look at the beauty of that result. We didn't need complex coordinate geometry or messy quadratic solutions. We used the properties of the circle, the symmetry of the triangle, and the elegance of trigonometry.

The final area is **$2\sqrt{2}$**. This is the essence of JEE Advanced mathematics—finding the path of least resistance through deep conceptual understanding.

Select Answer:

Visualized Solution

Visualizing the Circle

Grouping the $x$ and $y$ Terms

Completing the Square

Standard Form of the Circle

Identifying Center and Radius

Point $P$ and Line $C P$

The Diameter $QR$

Finding the Angles

Splitting Triangle $P QR$

Area Formula for $△ P C Q$

Calculating Area of $△ P C Q$

Area Formula for $△ P C R$

Calculating Area of $△ P C R$

Final Area Calculation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Anatomy of the Circle

The Geometric Setup

The Triangle Decomposition

The Final Calculation

Conclusion

Similar Questions

Let a triangle $A B C$ be inscribed in the circle $x^{2} - 2 (x + y) + y^{2} = 0$ such that $∠ B A C = \frac{π}{2}$ . If the length of side $A B$ is $2$ , then the area of the $△ A B C$ is equal to:

Let $C_{1}$ and $C_{2}$ be the centres of the circles $x^{2} + y^{2} - 2 x - 2 y - 2 = 0$ and $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$ respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $P C_{1} Q C_{2}$ is :

In the circle given below, let $O A = 1$ unit, $O B = 13$ unit and $P Q ∥ O B$ . Then, the area of the triangle PQB (in square units) is:

Let the line $x + y = 1$ meet the circle $x^{2} + y^{2} = 4$ at the points A and B. If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ADBC is equal to:

Let the parabola $y = x^{2} + p x - 3$ , meet the coordinate axes at the points P, Q and R. If the circle C with centre at $(- 1, - 1)$ passes through the points P, Q and R, then the area of $△ P QR$ is:

If the area of an equilateral triangle inscribed in the circle, $x^{2} + y^{2} + 10 x + 12 y + c = 0$ is $273$ sq. units then c is equal to :

The triangle $P QR$ is inscribed in the circle $x^{2} + y^{2} = 25$ . If $Q$ and $R$ have co-ordinates $(3, 4)$ and $(- 4, 3)$ respectively, then $∠ QP R$ is equal to

Let $C$ be a circle with radius $10$ units and centre at the origin. Let the line $x + y = 2$ intersects the circle $C$ at the points $P$ and $Q$ . Let $M N$ be a chord of $C$ of length 2 unit and slope -1. Then, a distance (in units) between the chord $P Q$ and the chord $M N$ is

Let the locus of the mid points of the chords of circle $x^{2} + (y - 1)^{2} = 1$ drawn from the origin intersect the line $x + y = 1$ at $P$ and $Q$ . Then, the length of $P Q$ is :

Select Answer:

Visualized Solution

Visualizing the Circle

Grouping the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y Terms

Completing the Square

Standard Form of the Circle

Identifying Center and Radius

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CP

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

Finding the Angles

Splitting Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQR

Area Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PCQ

Calculating Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PCQ

Area Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PCR

Calculating Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PCR