Let A be the point (1, 2) and B be any point on the curve x² + y² = 16. If the centre of the locus o

Q: Wait, why do we use parametric coordinates instead of just substituting $x^2 + y^2 = 16$ directly?

Great question! While you *could* use the Cartesian equation, it often leads to messy algebraic substitutions. Parametric coordinates ($4 \cos \theta, 4 \sin \theta$) are a powerful tool because they naturally satisfy the circle equation for any $\theta$. By using them, we reduce the problem to a simple trigonometric identity ($\cos^2 \theta + \sin^2 \theta = 1$), which is much cleaner and less prone to calculation errors.

Q: How do I know the locus is a circle? Is there a shortcut?

Yes, there is! This is a classic property in coordinate geometry. If you have a fixed point $A$ and a point $B$ moving on a circle, any point $P$ that divides the segment $AB$ in a fixed ratio $m:n$ will also trace out a circle. This is essentially a scaled and shifted version of the original circle. Recognizing this pattern can give you confidence in your result before you even finish the algebra.

Q: Why did we divide by 5 in the section formula? Is it always $m+n$?

Exactly! The internal section formula for a point $P$ dividing $AB$ in ratio $m:n$ is $P = \frac{mB + nA}{m+n}$. Here, the ratio is $3:2$, so $m=3$ and $n=2$. Thus, $m+n = 3+2 = 5$. Always remember to sum the ratio components in the denominator; it is a common place for silly mistakes!

Question

Let $A$ be the point $(1, 2)$ and $B$ be any point on the curve $x^2 + y^2 = 16$. If the centre of the locus of the point $P$, which divides the line segment $AB$ in the ratio $3 : 2$ is the point $C(\alpha, \beta)$, then the length of the line segment $AC$ is

Options:

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

(B) $\frac{4\sqrt{5}}{5}$

(C) $\frac{2\sqrt{5}}{5}$

(D) $\frac{6\sqrt{5}}{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You have a fixed anchor, point $A(1, 2)$, and a circular track defined by $x^2 + y^2 = 16$. This track is a perfect circle centered at the origin with a radius of $4$.

Now, imagine a point $B$ sliding along this track. As $B$ moves, a point $P$ is tethered to it, always maintaining a specific relationship: it divides the line segment $AB$ in a $3:2$ ratio. Our goal is to find the path—the locus—that $P$ traces as $B$ completes its journey around the circle.

### The Parametric Bridge

To capture the motion of $B$, we use the language of trigonometry. Since $B$ lies on a circle of radius $4$, we can describe its position using a single parameter, $	heta$. We define $B$ as $(4 \cos 	heta, 4 \sin 	heta)$.

We know that $P(h, k)$ divides $AB$ in the ratio $3:2$. Using the section formula, the coordinates of $P$ are the weighted average of the coordinates of $A$ and $B$:

$$h = \frac{3(4 \cos 	heta) + 2(1)}{3 + 2}, \quad k = \frac{3(4 \sin 	heta) + 2(2)}{3 + 2}$$

Simplifying these, we obtain the parametric equations for the locus:

$$h = \frac{12 \cos 	heta + 2}{5}, \quad k = \frac{12 \sin 	heta + 4}{5}$$

### Algebraic Alchemy

We now have $h$ and $k$ in terms of $	heta$, but we want the equation of the path of $P$ in terms of $x$ and $y$. We need to eliminate $	heta$. Rearranging our equations, we find:

$$\cos 	heta = \frac{5h - 2}{12}, \quad \sin 	heta = \frac{5k - 4}{12}$$

We invoke the fundamental identity $\cos^2 	heta + \sin^2 	heta = 1$. Substituting our expressions, we get:

$$\left( \frac{5h - 2}{12} ight)^2 + \left( \frac{5k - 4}{12} ight)^2 = 1$$

Multiplying by $12^2$, we arrive at the equation:

$$(5h - 2)^2 + (5k - 4)^2 = 144$$

### The Final Reveal

To see the true nature of this locus, we factor out the $5^2$ from the terms:

$$25\left(h - \frac{2}{5}ight)^2 + 25\left(k - \frac{4}{5}ight)^2 = 144$$

Dividing by $25$, we obtain the standard form of a circle:

$$\left(h - \frac{2}{5}ight)^2 + \left(k - \frac{4}{5}ight)^2 = \left(\frac{12}{5}ight)^2$$

The center $C$ of this locus is $(\frac{2}{5}, \frac{4}{5})$. The problem asks for the distance $AC$. With $A(1, 2)$ and $C(\frac{2}{5}, \frac{4}{5})$, we use the distance formula:

$$AC = \sqrt{\left(1 - \frac{2}{5}ight)^2 + \left(2 - \frac{4}{5}ight)^2}$$

This simplifies to:

$$AC = \sqrt{\left(\frac{3}{5}ight)^2 + \left(\frac{6}{5}ight)^2} = \sqrt{\frac{9}{25} + \frac{36}{25}} = \sqrt{\frac{45}{25}}$$

Simplifying the radical, the final distance is:

**$$AC = \frac{3\sqrt{5}}{5}$$**

Select Answer:

Visualized Solution

Visualizing the Setup

Parametric Coordinates of $B$

Applying Section Formula

Isolating Trigonometric Terms

Eliminating $θ$

Identifying the Center $C$

Calculating Distance $A C$

Final Result

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Parametric Bridge

Algebraic Alchemy

The Final Reveal

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