Let the circle x² + y² = 4 intersect x-axis at the points A(a, 0), a 0 and B(b, 0). Let P(2 cos α,

Q: Wait, why do we use half-angles? Can't we just use the coordinates directly?

You absolutely could, but it would lead to a nightmare of algebraic complexity! The beauty of using half-angles here is that the slope of a line connecting a point on the circle $(2\cos\theta, 2\sin\theta)$ to a point on the axis simplifies into a clean $\tan(\frac{\theta}{2})$ or $-\cot(\frac{\theta}{2})$. It transforms a trigonometric mess into a simple algebraic relationship, which is the hallmark of an elegant JEE solution.

Q: Why do we replace $(h, k)$ with $(x, y)$ at the end?

That is the standard procedure for finding a locus. We define the intersection point as $(h, k)$ to treat it as a specific, fixed point while we perform our calculations. Once we have the relationship between $h$ and $k$ that must hold true for any such intersection point, we replace them with the general variables $x$ and $y$ to describe the entire path or curve that the intersection point traces.

Q: What if the condition was $\alpha + \beta = \frac{\pi}{2}$ instead of $\alpha - \beta = \frac{\pi}{2}$?

That is a fantastic question! If the condition were different, the entire trigonometric identity we used would change. Instead of $\tan(\frac{\alpha}{2} - \frac{\beta}{2}) = 1$, you would be working with $\tan(\frac{\alpha}{2} + \frac{\beta}{2})$. This would lead to a completely different locus. The specific constraint given in the problem is what defines the path of the intersection point.

Question

Let the circle $x^2 + y^2 = 4$ intersect $x$-axis at the points $A(a, 0), a > 0$ and $B(b, 0)$. Let $P(2 \cos \alpha, 2 \sin \alpha)$, $0 < \alpha < \frac{\pi}{2}$ and $Q(2 \cos \beta, 2 \sin \beta)$ be two points such that $(\alpha - \beta) = \frac{\pi}{2}$. Then the point of intersection of $AQ$ and $BP$ lies on :

Options:

(A) $x^2 + y^2 - 4y - 4 = 0$

(B) $x^2 + y^2 - 4x - 4y = 0$

(C) $x^2 + y^2 - 4x - 4 = 0$

(D) $x^2 + y^2 - 4x - 4y - 4 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with a circle defined by the equation $x^2 + y^2 = 4$, which has a radius of $2$ and is centered at the origin. We are given two fixed points on the $x$-axis: $A(2, 0)$ and $B(-2, 0)$.

Two points, $P$ and $Q$, move along the circumference of this circle. Their coordinates are defined parametrically as $P(2\cos\alpha, 2\sin\alpha)$ and $Q(2\cos\beta, 2\sin\beta)$.

The motion is governed by the constraint $\alpha - \beta = \frac{\pi}{2}$. This constant phase difference ensures that as $P$ moves, $Q$ maintains a fixed relative position, creating a dynamic geometric system.

### The Slope Revelation

To find the intersection of lines $AQ$ and $BP$, we first determine their slopes. For line $BP$, the slope $m_{BP}$ is calculated as:

$$m_{BP} = \frac{2\sin\alpha - 0}{2\cos\alpha - (-2)} = \frac{\sin\alpha}{\cos\alpha + 1}$$

Using the half-angle identities $\sin\alpha = 2\sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2})$ and $1 + \cos\alpha = 2\cos^2(\frac{\alpha}{2})$, the expression simplifies to:

$$m_{BP} = 	an\left(\frac{\alpha}{2}ight)$$

Similarly, for line $AQ$, the slope $m_{AQ}$ is:

$$m_{AQ} = \frac{2\sin\beta - 0}{2\cos\beta - 2} = \frac{\sin\beta}{\cos\beta - 1}$$

Applying the identities $\sin\beta = 2\sin(\frac{\beta}{2})\cos(\frac{\beta}{2})$ and $\cos\beta - 1 = -2\sin^2(\frac{\beta}{2})$, we obtain:

$$m_{AQ} = -\cot\left(\frac{\beta}{2}ight)$$

### The Algebraic Bridge

We connect these slopes using the constraint $\alpha - \beta = \frac{\pi}{2}$, which implies $\frac{\alpha}{2} - \frac{\beta}{2} = \frac{\pi}{4}$. Taking the tangent of both sides, we apply the subtraction formula:

$$	an\left(\frac{\alpha}{2} - \frac{\beta}{2}ight) = \frac{	an(\frac{\alpha}{2}) - 	an(\frac{\beta}{2})}{1 + 	an(\frac{\alpha}{2})	an(\frac{\beta}{2})} = 1$$

This identity serves as the mathematical bridge linking the parameters of our two moving points.

### The Final Locus

Let the intersection point be $R(h, k)$. Since $R$ lies on $BP$ and $AQ$, we have $	an(\frac{\alpha}{2}) = \frac{k}{h+2}$ and $	an(\frac{\beta}{2}) = -\frac{h-2}{k}$.

Substituting these into our tangent identity, we get:

$$\frac{\frac{k}{h+2} - \left(-\frac{h-2}{k}ight)}{1 + \left(\frac{k}{h+2}ight)\left(-\frac{h-2}{k}ight)} = 1$$

Simplifying the numerator and denominator, the expression becomes:

$$\frac{k^2 + h^2 - 4}{4k} = 1$$

Rearranging the terms, we arrive at the equation $h^2 + k^2 - 4k - 4 = 0$. Replacing $(h, k)$ with $(x, y)$, the final locus is:

**$$x^2 + y^2 - 4y - 4 = 0$$**

Select Answer:

Visualized Solution

Equation of the Given Circle

Coordinates of Points $A$ and $B$

Parametric Coordinates of $P$ and $Q$

Slope of the Line $B P$

Simplifying $m_{B P}$ using Half-Angles

Slope of the Line $A Q$

Simplifying $m_{A Q}$ using Half-Angles

Applying the Angle Constraint

Intersection Point $R (h, k)$

Substituting into the Tangent Formula

Simplifying the Locus Equation

The Final Locus

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Slope Revelation

The Algebraic Bridge

The Final Locus

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Visualized Solution

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Coordinates of Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Parametric Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

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Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mAQ​ using Half-Angles

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