Let AB be a chord of the circle x² + y² = r² subtending a right angle at the centre. Then the locus

Q: Wait, why did we choose $A(r,0)$ and $B(0,r)$? Can't we choose any points?

You absolutely can! The beauty of this problem is its rotational symmetry. Choosing $A(r,0)$ and $B(0,r)$ makes the math incredibly clean, but the final locus will be the same regardless of where you place the chord, as long as it subtends $90^\circ$ at the center. It's a strategic choice, not a restriction.

Q: Why do we need to eliminate $\theta$?

$\theta$ is a parameter that describes the position of $P$. A locus equation must relate the coordinates $x$ and $y$ (or $h$ and $k$) directly. By eliminating $\theta$, we are essentially finding the 'path' that the centroid traces, independent of where $P$ is at any given moment.

Q: Is the locus always a circle?

In this specific case, yes! Because the centroid formula is a linear transformation of the coordinates of $P$, and $P$ moves on a circle, the centroid also moves on a circle. It's a scaled and shifted version of the original circle. Isn't that elegant?

Question

Let $AB$ be a chord of the circle $x^2 + y^2 = r^2$ subtending a right angle at the centre. Then the locus of the centroid of the triangle $PAB$ as $P$ moves on the circle is

Options:

(A) a parabola

(B) a circle

(C) an ellipse

(D) a pair of straight lines

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To solve this problem, we simplify our perspective by utilizing the symmetry of the circle. We place the fixed points $A$ and $B$ at coordinates $A(r, 0)$ and $B(0, r)$.

This configuration ensures that the chord $AB$ subtends a $90^{\circ}$ angle at the origin $(0,0)$, satisfying the problem's geometric constraints.

We define the moving point $P$ on the circle $x^2 + y^2 = r^2$ using parametric coordinates:
$$P(r\cos	heta, r\sin	heta)$$
Here, $	heta$ serves as the parameter that varies as $P$ traverses the circumference.

### The Heart of the Triangle

The centroid $G(h, k)$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is defined by the average of its coordinates:
$$h = \frac{x_1+x_2+x_3}{3}, \quad k = \frac{y_1+y_2+y_3}{3}$$

Applying this formula to our specific vertices $A(r, 0)$, $B(0, r)$, and $P(r\cos	heta, r\sin	heta)$, we obtain:
$$h = \frac{r + 0 + r\cos	heta}{3}$$
$$k = \frac{0 + r + r\sin	heta}{3}$$

### The Algebraic Dance

To find the locus of the centroid, we must eliminate the parameter $	heta$. We rearrange the equations to isolate the trigonometric terms:
$$3h - r = r\cos	heta$$
$$3k - r = r\sin	heta$$

We now utilize the fundamental trigonometric identity $\cos^2	heta + \sin^2	heta = 1$. By squaring both expressions and adding them, the parameter $	heta$ is eliminated:
$$(3h - r)^2 + (3k - r)^2 = (r\cos	heta)^2 + (r\sin	heta)^2$$

This simplifies to the following expression:
$$(3h - r)^2 + (3k - r)^2 = r^2(\cos^2	heta + \sin^2	heta) = r^2$$

### The Final Revelation

To reveal the geometric nature of this locus, we factor out the constant 3 from the squared terms:
$$9\left(h - \frac{r}{3}ight)^2 + 9\left(k - \frac{r}{3}ight)^2 = r^2$$

Dividing the entire equation by 9, we arrive at:
$$\left(h - \frac{r}{3}ight)^2 + \left(k - \frac{r}{3}ight)^2 = \left(\frac{r}{3}ight)^2$$

Replacing $(h, k)$ with the general coordinates $(x, y)$, we identify the final equation of the locus:
$$\left(x - \frac{r}{3}ight)^2 + \left(y - \frac{r}{3}ight)^2 = \left(\frac{r}{3}ight)^2$$

The centroid traces a **circle** centered at $(\frac{r}{3}, \frac{r}{3})$ with a radius of $\frac{r}{3}$.

Animated Solution for Mathematics - Circles: Let $A B$ be a chord of the circle $x^{2} + y^{2} = r^{2}$ subtending a right angle at the centre. Then the locus of the centroid of the triangle $P A B$ as $P$ moves on the circle is

Select Answer:

Visualized Solution

Visualizing the Circle

Defining Chord $A B$

The Moving Point $P$

Forming $△ P A B$

Centroid $G$ of $△ P A B$

Calculating the X-coordinate

Calculating the Y-coordinate

Isolating $cos θ$

Isolating $sin θ$

Trigonometric Identity

Eliminating $θ$

Simplifying the Equation

Standard Form of the Locus

Identifying the Locus

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Heart of the Triangle

The Algebraic Dance

The Final Revelation

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Select Answer:

Visualized Solution

Visualizing the Circle

Defining Chord .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

The Moving Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Forming .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PAB

Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PAB