If the chord y = mx + 1 of the circle x² + y² = 1 subtends an angle of measure 45° at the major segm

Q: Wait, why do we use the origin as the vertex for the angle?

The problem specifically asks for the angle subtended by the chord at the major segment of the circle. In coordinate geometry, when we talk about a chord subtending an angle at a point, if that point is the origin, we are essentially looking for the angle between the two lines that connect the origin to the endpoints of the chord. The origin is the natural reference point here.

Q: Why does homogenization work? Can't we just find the intersection points?

You absolutely could find the intersection points by solving the system of equations, but that would involve solving a quadratic equation, finding the coordinates, and then using the slope formula or the dot product to find the angle. It is algebraically tedious and prone to errors. Homogenization is a powerful, elegant shortcut that directly gives you the joint equation of the pair of lines, bypassing the need to find the individual intersection points.

Q: Why do we have a $\pm$ sign when we remove the modulus?

The formula $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$ involves an absolute value because the angle between two lines can be either acute or obtuse, and $\tan \theta$ changes sign depending on which angle you choose. When we set $\tan 45^\circ = 1$, we are essentially saying that the ratio must be $\pm 1$. This leads to two possible quadratic equations for $m$, which is why we get two possible values for the slope.

Question

If the chord $y = mx + 1$ of the circle $x^2 + y^2 = 1$ subtends an angle of measure $45^\circ$ at the major segment of the circle then value of $m$ is

Options:

(A) $2 \pm \sqrt{2}$

(B) $-2 \pm \sqrt{2}$

(C) $-1 \pm \sqrt{2}$

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the unit circle $x^2 + y^2 = 1$. It is a perfect, symmetric shape.

Now, consider a chord defined by the equation $y = mx + 1$. No matter what the slope $m$ is, this chord always passes through the point $(0, 1)$, which is the 'north pole' of our circle.

The problem asks us to find the slope $m$ such that this chord subtends an angle of $45^\circ$ at the origin. This is a classic JEE problem that tests your ability to choose the most elegant path to the solution.

### The Bridge: Homogenization

We could try to find the intersection points of the line and the circle, but that would lead us into a swamp of messy quadratic roots. Instead, we use a brilliant technique called **homogenization**.

We know the circle is $x^2 + y^2 = 1$. We can rewrite this as $x^2 + y^2 = 1^2$. The chord equation is $y = mx + 1$, which we rearrange to $y - mx = 1$.

By substituting this '1' into our circle equation, we transform the linear constraint of the chord into a quadratic form that matches the circle:

$$x^2 + y^2 = (y - mx)^2$$

This is the magic of homogenization: it creates a second-degree equation that represents the pair of lines connecting the origin to the intersection points of the chord and the circle.

### The Algebraic Dance

Now, let us expand the right side:

$$x^2 + y^2 = y^2 - 2mxy + m^2x^2$$

Notice the beauty of the cancellation! The $y^2$ terms on both sides vanish, leaving us with:

$$x^2 = -2mxy + m^2x^2$$

Rearranging this into the standard form for a pair of lines, $ax^2 + 2hxy + by^2 = 0$, we get:

$$x^2(1 - m^2) + 2mxy + 0 \cdot y^2 = 0$$

Here, our coefficients are $a = 1 - m^2$, $h = m$, and $b = 0$. This equation represents the two lines originating from $(0, 0)$ that pass through the endpoints of our chord.

### The Final Calculation

We know the angle $	heta$ between two lines $ax^2 + 2hxy + by^2 = 0$ is given by:

$$	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$$

Given $	heta = 45^\circ$, we have $	an 45^\circ = 1$. Substituting our values, we get:

$$1 = \left| \frac{2\sqrt{m^2 - (1 - m^2)(0)}}{(1 - m^2) + 0} ight|$$

This simplifies to:

$$1 = \left| \frac{2m}{1 - m^2} ight|$$

Removing the modulus gives us $1 - m^2 = \pm 2m$. This leads to two quadratic equations:

$$m^2 + 2m - 1 = 0 \quad 	ext{and} \quad m^2 - 2m - 1 = 0$$

Solving these using the quadratic formula, we find:

$$m = -1 \pm \sqrt{2} \quad 	ext{and} \quad m = 1 \pm \sqrt{2}$$

The problem asks for the value of $m$. By using homogenization, we turned a potentially grueling problem into a clean, elegant algebraic exercise. The final values for the slope are **$m = -1 \pm \sqrt{2}$** or **$m = 1 \pm \sqrt{2}$**.

Animated Solution for Mathematics - Circles: If the chord $y = m x + 1$ of the circle $x^{2} + y^{2} = 1$ subtends an angle of measure $4 5^{\circ}$ at the major segment of the circle then value of $m$ is

Select Answer:

Visualized Solution

Geometry of the Circle and Chord

The Homogenization Concept

Applying Homogenization

Expanding the Equation

Standard Form of Pair of Lines

Angle Between Pair of Lines

Substituting the Values

Solving the Modulus Equation

Applying the Quadratic Formula

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Bridge

The Algebraic Dance

The Final Calculation

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