Let 2x² + y² - 3xy = 0 be the equation of a pair of tangents drawn from the origin O to a circle of

Q: Why do we need to factorize the equation $2x^2 - 3xy + y^2 = 0$?

The equation $2x^2 - 3xy + y^2 = 0$ represents a pair of lines passing through the origin. To perform any geometric analysis—like finding the angle between them—we need the individual equations of these lines. Factorization allows us to isolate the two linear equations, $y=2x$ and $y=x$, giving us the slopes $m_1=2$ and $m_2=1$ necessary for our calculations.

Q: Why is the angle at the origin $\frac{\theta}{2}$?

This is a fundamental property of tangents drawn from an external point to a circle. The line segment connecting the external point (the origin) to the center of the circle acts as the angle bisector of the two tangents. Therefore, if the total angle between the tangents is $\theta$, the angle between one tangent and the line connecting the origin to the center is exactly half of that, or $\frac{\theta}{2}$.

Q: Why do we reject the negative root in the quadratic equation $t^2 + 6t - 1 = 0$?

We defined $t = \tan(\frac{\theta}{2})$. Since the tangents lie in the first quadrant, the angle $\theta$ between them is acute, meaning $\frac{\theta}{2}$ is also acute. The tangent of an acute angle must be positive. The roots of our quadratic are $-3 \pm \sqrt{10}$. Since $\sqrt{10} \approx 3.16$, the root $-3 - \sqrt{10}$ is negative, which is physically impossible for our geometry. Thus, we must choose the positive root, $\sqrt{10} - 3$.

Question

Let $2x^2 + y^2 - 3xy = 0$ be the equation of a pair of tangents drawn from the origin $O$ to a circle of radius 3 with centre in the first quadrant. If $A$ is one of the points of contact, find the length of $OA$.

Sigma Percentile · Accepted Answer

### The Geometry of the Origin: A Journey into Tangents

Welcome, student. Today, we are not just solving a coordinate geometry problem; we are peeling back the layers of a beautiful geometric structure.

Imagine you are standing at the origin of a Cartesian plane. You look out into the first quadrant, and you see two lines emanating from where you stand. These lines are not random; they are the guardians of a circle, a perfect circle of radius $r=3$, nestled between them.

Our mission is to find the length of the tangent segment from the origin to the point of contact on this circle. Let us begin.

### Phase 1: Deconstructing the Pair of Lines

We are given the equation $2x^2 - 3xy + y^2 = 0$. This is a second-degree homogeneous equation, a classic signature of a pair of straight lines passing through the origin.

To work with them, we factorize this expression by splitting the middle term:
$$2x^2 - 2xy - xy + y^2 = 0$$

This simplifies beautifully to:
$$(2x - y)(x - y) = 0$$

Just like that, the fog clears. We have two distinct lines: $y = 2x$ and $y = x$. Their slopes are $m_1 = 2$ and $m_2 = 1$. These are the two paths that define the boundaries of our circle's existence.

### Phase 2: The Angle of Symmetry

Now, we need to understand the relationship between these two lines. Let $	heta$ be the angle between them.

We use the standard formula for the angle between two lines:
$$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$$

Substituting our slopes, we get:
$$	an 	heta = \left| \frac{2 - 1}{1 + 2(1)} ight| = \frac{1}{3}$$

This value, $\frac{1}{3}$, is the key to the entire problem. The center of the circle, $C$, must lie on the angle bisector of these two lines.

If we draw a line from the origin $O$ to the center $C$, it bisects the angle $	heta$. Thus, the angle $\angle AOC$ is exactly $\frac{	heta}{2}$.

### Phase 3: The Right-Angled Bridge

Let $A$ be the point of contact on one of the tangents. We know that the radius $CA$ is perpendicular to the tangent $OA$.

This creates a right-angled triangle, $	riangle OAC$, where $\angle OAC = 90^\circ$. In this triangle, we have the radius $CA = 3$ and the angle $\angle AOC = \frac{	heta}{2}$.

Using basic trigonometry, we define the relationship:
$$	an\left(\frac{	heta}{2}ight) = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{CA}{OA}$$

Rearranging this, we find the length we seek:
$$OA = 3 \cot\left(\frac{	heta}{2}ight)$$

### Phase 4: The Half-Angle Transformation

We know $	an 	heta = \frac{1}{3}$, but we need $\cot(\frac{	heta}{2})$. We use the half-angle identity:
$$	an 	heta = \frac{2 	an(\frac{	heta}{2})}{1 - 	an^2(\frac{	heta}{2})}$$

Let $t = 	an(\frac{	heta}{2})$. Our equation becomes:
$$\frac{1}{3} = \frac{2t}{1 - t^2}$$

Cross-multiplying gives us $1 - t^2 = 6t$, or $t^2 + 6t - 1 = 0$. Solving this quadratic equation using the quadratic formula:
$$t = \frac{-6 \pm \sqrt{36 - 4(1)(-1)}}{2} = -3 \pm \sqrt{10}$$

Since $\frac{	heta}{2}$ is an acute angle, its tangent must be positive. Therefore, we reject the negative root and accept $t = \sqrt{10} - 3$.

### The Final Victory

We need $\cot(\frac{	heta}{2})$, which is:
$$\cot\left(\frac{	heta}{2}ight) = \frac{1}{t} = \frac{1}{\sqrt{10} - 3}$$

Rationalizing the denominator by multiplying by the conjugate $\sqrt{10} + 3$, we get:
$$\frac{\sqrt{10} + 3}{10 - 9} = \sqrt{10} + 3$$

Finally, we return to our expression for the tangent length:
$$OA = 3 \cot\left(\frac{	heta}{2}ight) = 3(\sqrt{10} + 3) = 9 + 3\sqrt{10}$$

There it is. The length of the tangent $OA$ is **$9 + 3\sqrt{10}$**. You have successfully navigated the factorization, the geometric symmetry, the trigonometric identities, and the algebraic resolution.

Animated Solution for Mathematics - Circles: Let $2 x^{2} + y^{2} - 3 x y = 0$ be the equation of a pair of tangents drawn from the origin $O$ to a circle of radius 3 with centre in the first quadrant. If $A$ is one of the points of contact, find the length of $O A$ .

Enter Numerical Value:

Visualized Solution

Pair of Tangents Equation

Factorizing to Find Slopes

Angle Between Tangents

Circle and Angle Bisector

Right Triangle $△ O A C$

Relating $O A$ and $θ /2$

Using the Half-Angle Formula

Forming a Quadratic in $t$

Solving the Quadratic Equation

Finding $cot (θ /2)$

Final Length of Tangent $O A$

The Sigma Insight: Equation of Tangent and Normal

The Geometry of the Origin

Phase 1

Phase 2

Phase 3

Phase 4

The Final Victory

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