Let the tangents at two points A and B on the circle x² + y² - 4x + 3 = 0 meet at origin O(0, 0). Th

Q: Wait, why can't I just find the coordinates of points A and B by solving the intersection of the tangent lines and the circle?

You absolutely could! But in the JEE, time is your most precious currency. Solving for the intersection of a line and a circle involves solving a quadratic equation, which is prone to calculation errors. Using the geometric properties of the tangent—specifically the right triangle $\triangle OAC$—allows you to bypass the algebra and reach the answer with elegant precision.

Q: How do I know for sure that $\angle AOB = 2 \times \angle AOC$?

Great question! This comes from the symmetry of the tangents drawn from an external point to a circle. The line segment connecting the external point $O$ to the center $C$ acts as an angle bisector for the angle formed by the two tangents. Thus, $\triangle OAC$ and $\triangle OBC$ are congruent right-angled triangles, confirming the symmetry.

Q: Is the length of the tangent always $\sqrt{S_1}$?

Yes, for any circle $S: x^2 + y^2 + 2gx + 2fy + c = 0$, the length of the tangent from a point $(x_1, y_1)$ is indeed $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$. It is a powerful shortcut derived from the Pythagorean theorem applied to the triangle formed by the center, the point of tangency, and the external point.

Q: Why did we use the area formula $\frac{1}{2}ab \sin(C)$ instead of $\frac{1}{2} \times \text{base} \times \text{height}$?

You can use $\frac{1}{2} \times \text{base} \times \text{height}$, but you would first need to find the length of the chord of contact $AB$ and the altitude from $O$ to $AB$. The $\frac{1}{2}OA \cdot OB \cdot \sin(\angle AOB)$ formula is a direct application of the SAS (Side-Angle-Side) theorem, which is much faster when you already know the lengths of the tangents and the angle between them.

Q: What if the origin was inside the circle? Would the tangent length formula still work?

If the origin were inside the circle, the value of $S_1$ would be negative, making $\sqrt{S_1}$ an imaginary number. This is the math's way of telling you that it is physically impossible to draw a tangent to a circle from a point inside it! Always check your point's position relative to the circle first.

Question

Let the tangents at two points $A$ and $B$ on the circle $x^2 + y^2 - 4x + 3 = 0$ meet at origin $O(0, 0)$. Then the area of the triangle of $OAB$ is

Options:

(A) $\frac{3\sqrt{3}}{2}$

(B) $\frac{3\sqrt{3}}{4}$

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

(D) $\frac{3}{4\sqrt{3}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane, looking out at a circle defined by the equation $x^2 + y^2 - 4x + 3 = 0$. To the untrained eye, this is just a collection of algebraic terms.

By completing the square, we rewrite the equation as $(x - 2)^2 + y^2 = 1^2$. Suddenly, the fog clears: we are looking at a circle centered at $C(2, 0)$ with a radius $r = 1$.

### The Power of Symmetry

We are tasked with finding the area of the triangle $	riangle OAB$, where $OA$ and $OB$ are tangents drawn from the origin $O(0, 0)$ to the circle. Many students immediately reach for the quadratic formula to find the coordinates of $A$ and $B$.

I urge you to pause. In the JEE, the most beautiful path is rarely the most brute-forced one. Instead, let us look at the geometry.

Consider the right-angled triangle $	riangle OAC$. Here, $OA$ is the tangent, $AC$ is the radius, and $OC$ is the distance from the origin to the center. Since the radius is always perpendicular to the tangent at the point of contact, $\angle OAC = 90^{\circ}$.

We know the length of the tangent $OA$ from the origin is given by $\sqrt{S_1}$, where $S_1$ is the value of the circle's equation at the origin. Substituting $(0, 0)$ into $x^2 + y^2 - 4x + 3$, we get the length $L$:

$$L = \sqrt{3}$$

### The Trigonometric Bridge

Now, look at the triangle $	riangle OAC$ again. We have the side $AC = 1$ (the radius) and the hypotenuse $OC = 2$ (the distance from $(0, 0)$ to $(2, 0)$).

The sine of the angle $	heta = \angle AOC$ is simply the ratio of the opposite side to the hypotenuse:

$$\sin(	heta) = \frac{AC}{OC} = \frac{1}{2}$$

This tells us that $	heta = 30^{\circ}$. Because the two tangents are symmetric with respect to the line $OC$, the total angle $\angle AOB$ is simply $2	heta = 60^{\circ}$.

We have effectively reduced a complex coordinate geometry problem into a simple triangle with two sides of length $\sqrt{3}$ and an included angle of $60^{\circ}$.

### The Final Elegance

To find the area of $	riangle OAB$, we use the SAS area formula, which is the most efficient tool in our kit:

$$	ext{Area} = \frac{1}{2} \cdot OA \cdot OB \cdot \sin(\angle AOB)$$

Substituting our known values:

$$	ext{Area} = \frac{1}{2} \cdot (\sqrt{3}) \cdot (\sqrt{3}) \cdot \sin(60^{\circ})$$

Since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$, the calculation becomes a satisfyingly simple arithmetic exercise:

$$	ext{Area} = \frac{1}{2} \cdot 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4}$$

Look at that result. It is clean, precise, and derived not through tedious algebra, but through a deep understanding of the geometric relationships between the circle and the tangents.

The **final answer** is **$\frac{3\sqrt{3}}{4}$**. Remember, the JEE does not just test your ability to calculate; it tests your ability to see the underlying structure of the problem.

Animated Solution for Mathematics - Circles: Let the tangents at two points $A$ and $B$ on the circle $x^{2} + y^{2} - 4 x + 3 = 0$ meet at origin $O (0, 0)$ . Then the area of the triangle of $O A B$ is

Select Answer:

Visualized Solution

The Given Circle

Completing the Square

Center and Radius

Tangents from Origin

Length of Tangent Formula

Substituting the Origin

Calculating Tangent Length

Constructing $△ O A C$

Analyzing $△ O A C$

Finding Angle $∠ A O C$

Total Angle $∠ A O B$

Area of $△ O A B$ Formula

Substituting Values for Area

Final Calculation

Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Power of Symmetry

The Trigonometric Bridge

The Final Elegance

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