If the tangents at the points P and Q on the circle x² + y² - 2x + y = 5 meet at the point R(9/4, 2)

Q: Wait, why can't we just find the coordinates of $P$ and $Q$ and use the area of a triangle formula? It seems more direct.

While finding the coordinates of $P$ and $Q$ is mathematically possible, it is a trap! You would have to solve the intersection of the tangent lines with the circle, leading to messy quadratic equations and potential calculation errors. In JEE Advanced, time is your most precious resource. Using the derived formula $\text{Area} = \frac{r L^3}{r^2 + L^2}$ is the 'pro' way to bypass the algebra and reach the answer with elegance.

Q: How do I know for sure that the point $R$ is outside the circle?

Excellent question! Always check the power of the point. If you substitute the coordinates of $R$ into the circle equation $S(x, y) = x^2 + y^2 - 2x + y - 5$, and the result is positive, the point is outside. If it were zero, the point would be on the circle, and the tangent length would be zero. If negative, it would be inside, and no real tangents could be drawn. Here, $S_1 = \frac{25}{16} > 0$, confirming $R$ is safely outside.

Q: Is the formula $\text{Area} = \frac{r L^3}{r^2 + L^2}$ valid for all circles?

Yes, it is a universal geometric result for any circle of radius $r$ and any external point from which tangents of length $L$ are drawn. It stems from the similarity of triangles within the configuration. Once you derive or memorize this, you unlock a powerful shortcut that saves you minutes of tedious coordinate geometry.

Question

If the tangents at the points $P$ and $Q$ on the circle $x^2 + y^2 - 2x + y = 5$ meet at the point $R(\frac{9}{4}, 2)$, then the area of the triangle $PQR$ is

Options:

(A) $\frac{5}{4}$

(B) $\frac{13}{8}$

(C) $\frac{5}{8}$

(D) $\frac{13}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, students! Today, we are going to explore a classic JEE Advanced problem that tests not just your algebraic stamina, but your geometric intuition.

We are given a circle defined by $x^2 + y^2 - 2x + y - 5 = 0$ and an external point $R(\frac{9}{4}, 2)$. Our mission is to find the area of the triangle $PQR$, where $P$ and $Q$ are the points of tangency from $R$.

The triangle formed by these tangents and the chord of contact $PQ$ is a symmetric structure. Let's break this down into a journey of discovery.

### The Circle's Identity

Before we can do anything, we must understand our circle. The equation $x^2 + y^2 - 2x + y - 5 = 0$ is in the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.

By comparing coefficients, we find $2g = -2$, so $g = -1$, and $2f = 1$, so $f = \frac{1}{2}$. The center $C$ is $(-g, -f)$, which gives us $C(1, -\frac{1}{2})$.

Now, for the radius $r$, we use the formula $r = \sqrt{g^2 + f^2 - c}$. Substituting our values:

$$r = \sqrt{(-1)^2 + \left(\frac{1}{2}ight)^2 - (-5)} = \sqrt{1 + \frac{1}{4} + 5} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$

We have a circle with center $(1, -\frac{1}{2})$ and radius $r = \frac{5}{2}$.

### The Tangent's Power

Now, let's look at the external point $R(\frac{9}{4}, 2)$. We need the length of the tangents $RP$ and $RQ$. Let this length be $L$.

The power of a point $R$ with respect to the circle is $S_1$, and the length of the tangent is simply $L = \sqrt{S_1}$. Substituting $R(\frac{9}{4}, 2)$ into the circle equation $S(x, y) = x^2 + y^2 - 2x + y - 5$:

$$S_1 = \left(\frac{9}{4}ight)^2 + (2)^2 - 2\left(\frac{9}{4}ight) + 2 - 5$$

$$S_1 = \frac{81}{16} + 4 - \frac{9}{2} + 2 - 5 = \frac{81}{16} + 1 - \frac{72}{16} = \frac{9}{16} + 1 = \frac{25}{16}$$

Thus, $L = \sqrt{\frac{25}{16}} = \frac{5}{4}$. We have found the length of our tangents.

### The Geometric Elegance

We use the standard JEE result for the area of the triangle formed by two tangents of length $L$ and the chord of contact in a circle of radius $r$:

$$	ext{Area} = \frac{r L^3}{r^2 + L^2}$$

Look at our values: $r = \frac{5}{2}$ and $L = \frac{5}{4}$. Notice that $r = 2L$. Let's substitute $r = 2L$ into our formula:

$$	ext{Area} = \frac{(2L) L^3}{(2L)^2 + L^2} = \frac{2L^4}{4L^2 + L^2} = \frac{2L^4}{5L^2} = \frac{2}{5} L^2$$

### Final Calculation

Now, the final calculation is trivial:

$$	ext{Area} = \frac{2}{5} 	imes \left(\frac{5}{4}ight)^2 = \frac{2}{5} 	imes \frac{25}{16} = \frac{5}{8}$$

The elegance of this cancellation is why we love mathematics. We have arrived at the final answer of **$\frac{5}{8}$** with precision and grace.

Animated Solution for Mathematics - Circles: If the tangents at the points $P$ and $Q$ on the circle $x^{2} + y^{2} - 2 x + y = 5$ meet at the point $R (\frac{9}{4}, 2)$ , then the area of the triangle $P QR$ is

Select Answer:

Visualized Solution

Visualizing the Problem

Finding the Center of the Circle

Setting up the Radius Formula

Calculating the Radius $r$

Drawing the Tangents

Length of Tangent Formula

Substituting Point $R$ into $S_{1}$

Calculating Tangent Length $L$

Forming Triangle $P QR$

Area Formula for $△ P QR$

Substituting $r$ and $L$

Simplifying the Area Expression

Final Area Calculation

The Sigma Insight: Length of Tangent and Chord of Contact

Analyzing the Setup

The Circle's Identity

The Tangent's Power

The Geometric Elegance

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Problem

Finding the Center of the Circle

Setting up the Radius Formula

Calculating the Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Drawing the Tangents

Length of Tangent Formula

Substituting Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Calculating Tangent Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Forming 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; } △PQR

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L