Let x² + y² - 4x - 2y - 11 = 0 be a circle. A pair of tangents from the point (4, 5) with a pair of

Q: Why is the tangent length formula $L = \sqrt{S_1}$? Can't I just use the distance formula?

Great question! The distance formula gives you the distance between two points. Here, we need the length of the tangent segment. The formula $L = \sqrt{S_1}$ is actually a direct consequence of the Pythagorean theorem applied to the triangle formed by the center, the point of contact, and the external point. It saves you the trouble of finding the exact point of contact!

Q: Why do we split the quadrilateral into two triangles? Is there a direct formula for the area?

You could memorize a formula, but in JEE, visualization is your best friend. Splitting the quadrilateral into two congruent right-angled triangles is the most robust method. It reduces the problem to a simple $\frac{1}{2} \times \text{base} \times \text{height}$ calculation, minimizing the chance of algebraic errors.

Q: How do I know the two triangles are congruent?

Excellent observation! The two triangles share the hypotenuse (the line segment connecting the external point to the center). They both have a side equal to the radius of the circle. By the RHS (Right-angle, Hypotenuse, Side) congruence criterion, they are identical. This symmetry is the beauty of circle geometry!

Question

Let $x^2 + y^2 - 4x - 2y - 11 = 0$ be a circle. A pair of tangents from the point $(4, 5)$ with a pair of radii form a quadrilateral of area .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a coordinate geometry problem; we are uncovering the hidden symmetry within a circle.

Imagine you are standing at point $P(4, 5)$, looking down at a circle defined by $x^2 + y^2 - 4x - 2y - 11 = 0$. You draw two lines from your position that just barely graze the circle—these are your tangents.

Connect the points where these tangents touch the circle to the center, and you have created a quadrilateral. It looks simple, but it holds a beautiful secret.

### Decoding the Circle

Before we can dance with the geometry, we must know our partner. The equation $x^2 + y^2 - 4x - 2y - 11 = 0$ is our starting point.

To find the heart of the circle—its center—and its size—its radius—we compare this to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$. By matching coefficients, we find $2g = -4$ and $2f = -2$, leading us to the center $C(2, 1)$.

The radius, calculated via $r = \sqrt{g^2 + f^2 - c}$, is determined as follows:

$$r = \sqrt{(-2)^2 + (-1)^2 - (-11)} = \sqrt{4 + 1 + 11} = \sqrt{16} = 4$$

We now have a circle centered at $C(2, 1)$ with a radius of $4$.

### The Tangent Mystery

Now, we turn our attention to the tangents from $P(4, 5)$. We need their length.

While we could find the points of contact using complex intersection algebra, there is a more elegant path. The length of a tangent $L$ from an external point $(x_1, y_1)$ is simply $\sqrt{S_1}$, where $S_1$ is the value of the circle's equation at that point.

Substituting $x=4$ and $y=5$ into our equation, we get:

$$S_1 = (4)^2 + (5)^2 - 4(4) - 2(5) - 11$$

$$S_1 = 16 + 25 - 16 - 10 - 11 = 4$$

Thus, the length of our tangent $L = \sqrt{4} = 2$.

### The Geometric Insight

Here is where the magic happens. We have a quadrilateral formed by the two tangents and the two radii.

The radius is always perpendicular to the tangent at the point of contact. This means we have two right-angled triangles, $	riangle PT_1C$ and $	riangle PT_2C$.

Because they share the same hypotenuse (the line $PC$) and have the same radius $r$, they are perfectly congruent. We don't need to calculate the area of the quadrilateral all at once; we just need to find the area of one triangle and double it.

### Final Calculation

For our right-angled triangle $	riangle PT_1C$, the base is the tangent length $L=2$, and the height is the radius $r=4$.

The area of this triangle is calculated as:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 4 = 4$$

Since the quadrilateral is composed of two such triangles, the total area is $2 	imes 4 = 8$ square units.

It is elegant, it is precise, and it is the power of geometric intuition. The final area is **8 square units**.

Animated Solution for Mathematics - Circles: Let $x^{2} + y^{2} - 4 x - 2 y - 11 = 0$ be a circle. A pair of tangents from the point $(4, 5)$ with a pair of radii form a quadrilateral of area .........

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Extracting Circle Parameters

Calculating the Center

Calculating the Radius

Length of Tangent Formula

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

Evaluating Tangent Length

Forming the Quadrilateral

The $9 0^{\circ}$ Tangent-Radius Theorem

Area of One Right Triangle

Computing Triangle Area

Total Area of Quadrilateral

The Sigma Insight: Length of Tangent and Chord of Contact

Analyzing the Setup

Decoding the Circle

The Tangent Mystery

The Geometric Insight

Final Calculation

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