A line segment AB of length λ moves such that the points A and B remain on the periphery of a circle

Q: Wait, why is the triangle $OAB$ equilateral? I thought it was just a random chord.

That is a brilliant observation! In any circle, if the length of a chord is equal to the radius, the triangle formed by the two endpoints and the center has all three sides equal to the radius. Since all sides are $\lambda$, it must be equilateral. This is the 'hidden' geometric key that unlocks the problem!

Q: Why do we use parametric coordinates instead of just Cartesian coordinates?

Great question. Cartesian coordinates $(x, y)$ are static, but the segment $AB$ is moving. By using $\theta_1$ and $\theta_2$, we capture the 'motion' of the segment. It turns the geometry problem into an algebraic one, allowing us to use trigonometric identities to find the path.

Q: Is it always true that $h^2 + k^2$ gives the radius squared?

Yes, exactly! The equation $h^2 + k^2 = R^2$ represents a circle centered at the origin with radius $R$. By manipulating our section formula results into this form, we are essentially proving that the point $P$ traces a circular path.

Q: Why did we use the cosine subtraction formula? Where did that come from?

When we expanded $(3 \cos \theta_2 + 2 \cos \theta_1)^2 + (3 \sin \theta_2 + 2 \sin \theta_1)^2$, the cross-terms were $12(\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2)$. That expression is the classic expansion for $\cos(\theta_1 - \theta_2)$. It is a beautiful way to collapse complex trigonometry into a single constant value.

Q: What if the ratio was different, say $1:1$? Would the radius change?

Absolutely! The ratio determines the 'weight' of each point in the section formula. If the ratio changes, the coefficients in our expansion change, which directly alters the final radius. The beauty of this method is that it is a general template for any internal division ratio.

Question

A line segment $AB$ of length $\lambda$ moves such that the points $A$ and $B$ remain on the periphery of a circle of radius $\lambda$. Then the locus of the point, that divides the line segment $AB$ in the ratio $2 : 3$, is a circle of radius

Options:

(A) $\frac{3}{5}\lambda$

(B) $\frac{2}{3}\lambda$

(C) $\frac{\sqrt{19}}{5}\lambda$

(D) $\frac{\sqrt{19}}{7}\lambda$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty plane. Before you lies a circle of radius $\lambda$, centered perfectly at the origin.

You hold a rod of length $\lambda$, and you place its ends, $A$ and $B$, on the boundary of this circle. As you slide the rod around, the points $A$ and $B$ dance along the circumference.

Your task is to track a specific point $P$ on this rod—a point that divides the segment $AB$ in a $2:3$ ratio. We seek to determine the path followed by this point.

### The Equilateral Insight

Before we dive into the algebra, let us pause and appreciate the geometry. We are told the length of the chord $AB$ is $\lambda$.

Since the distance from the center $O$ to any point on the circle is also $\lambda$, the triangle $	riangle OAB$ has sides of length $\lambda, \lambda,$ and $\lambda$. It is an equilateral triangle!

This is our first major breakthrough. It tells us that the angle subtended by the chord at the center is fixed at $60^{\circ}$, or $\frac{\pi}{3}$ radians. This constant angle is the anchor for our entire derivation.

### Parametric Elegance

To track the motion, we assign coordinates to $A$ and $B$. Let $A = (\lambda \cos 	heta_2, \lambda \sin 	heta_2)$ and $B = (\lambda \cos 	heta_1, \lambda \sin 	heta_1)$.

Because the rod is rigid and the triangle is equilateral, the difference between these angles must be constant: $|	heta_1 - 	heta_2| = \frac{\pi}{3}$.

Now, we invoke the section formula. For a point $P(h, k)$ dividing $AB$ in a $2:3$ ratio, we have:

$$P = \frac{3A + 2B}{3+2}$$

This gives us the coordinates of our point $P$ at any moment in time:

$$h = \frac{\lambda}{5} (3 \cos 	heta_2 + 2 \cos 	heta_1)$$

$$k = \frac{\lambda}{5} (3 \sin 	heta_2 + 2 \sin 	heta_1)$$

### The Algebraic Dance

We want the locus of $P$. To find it, we need to eliminate the parameters $	heta_1$ and $	heta_2$.

The most powerful tool in our arsenal is the identity $\cos^2 	heta + \sin^2 	heta = 1$. Let us square $h$ and $k$ and add them:

$$h^2 + k^2 = \frac{\lambda^2}{25} [ (3 \cos 	heta_2 + 2 \cos 	heta_1)^2 + (3 \sin 	heta_2 + 2 \sin 	heta_1)^2 ]$$

When we expand these squares, the magic happens. The terms $9(\cos^2 	heta_2 + \sin^2 	heta_2)$ and $4(\cos^2 	heta_1 + \sin^2 	heta_1)$ simplify beautifully to $9(1)$ and $4(1)$.

The cross-terms combine into $12(\cos 	heta_1 \cos 	heta_2 + \sin 	heta_1 \sin 	heta_2)$. Recognizing this as the cosine subtraction formula, we get:

$$h^2 + k^2 = \frac{\lambda^2}{25} [ 9 + 4 + 12 \cos(	heta_1 - 	heta_2) ]$$

### The Final Revelation

We know that $\cos(	heta_1 - 	heta_2) = \cos(60^{\circ}) = \frac{1}{2}$. Substituting this into our equation, the complexity collapses:

$$h^2 + k^2 = \frac{\lambda^2}{25} [ 13 + 12(\frac{1}{2}) ] = \frac{\lambda^2}{25} [ 13 + 6 ] = \frac{19\lambda^2}{25}$$

This is the equation of a circle! Specifically, $x^2 + y^2 = (\frac{\sqrt{19}}{5}\lambda)^2$.

The point $P$ is not wandering; it is tracing a perfect circle of radius **$\frac{\sqrt{19}}{5}\lambda$**. You have successfully tamed the motion and revealed the hidden symmetry of the system.

Animated Solution for Mathematics - Circles: A line segment $A B$ of length $λ$ moves such that the points $A$ and $B$ remain on the periphery of a circle of radius $λ$ . Then the locus of the point, that divides the line segment $A B$ in the ratio $2 : 3$ , is a circle of radius

Select Answer:

Visualized Solution

Visualize the Circle and Chord $A B$

The Equilateral Triangle $△ O A B$

Parametric Coordinates of $A$ and $B$

Locating Point $P (h, k)$

Applying Section Formula for $h$

Applying Section Formula for $k$

Strategy: Square and Add

Squaring the Equations

Expanding the Squares

Using Trigonometric Identities

Substituting the Angle Difference

Atomic Compute: Final Simplification

The Radius of the Locus

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Equilateral Insight

Parametric Elegance

The Algebraic Dance

The Final Revelation

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