A circle of radius 'r' passes through the origin `O` and cuts the axes at A and B,Locus of the centroid of triangle OAB is

Options

A. `(x^(2)+y^(2))^(2)=4Rx^(2)y^(2)`
B. `(x^(2)+y^(2))(x+y)=R^(2)xy`
C. `(x^(2)+y^(2))^(3)=4R^(2)x^(2)y^(2)`
D. `(x^(2)+y^(2))^(2)=4R^(2)x^(2)y^(2)`

Answer

To find the locus of the centroid of triangle OAB formed by a circle of radius 'r' that passes through the origin and intersects the axes at points A and B, we can follow these steps:

### Step 1: Define the Circle
The equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Since the circle passes through the origin (0, 0), we can substitute these coordinates into the equation:
\[ (0 - h)^2 + (0 - k)^2 = r^2 \implies h^2 + k^2 = r^2 \]

### Step 2: Identify Points A and B
Let the circle intersect the x-axis at point A and the y-axis at point B. The coordinates of these points can be defined as:
- Point A: \( (a, 0) \)
- Point B: \( (0, b) \)

### Step 3: Use the Circle's Equation
Since points A and B lie on the circle, we can substitute these points into the circle's equation:
1. For point A:
\[ (a - h)^2 + (0 - k)^2 = r^2 \]
This simplifies to:
\[ (a - h)^2 + k^2 = r^2 \]

2. For point B:
\[ (0 - h)^2 + (b - k)^2 = r^2 \]
This simplifies to:
\[ h^2 + (b - k)^2 = r^2 \]

### Step 4: Calculate the Centroid
The centroid \( G \) of triangle OAB is given by:
\[ G\left(\frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}\right) = \left(\frac{a}{3}, \frac{b}{3}\right) \]
Let \( G = (h, k) \) where \( h = \frac{a}{3} \) and \( k = \frac{b}{3} \).

### Step 5: Express a and b in terms of h and k
From the centroid coordinates, we can express:
\[ a = 3h \quad \text{and} \quad b = 3k \]

### Step 6: Substitute a and b into the Circle's Equation
Substituting \( a \) and \( b \) into the circle's equations:
1. From point A:
\[ (3h - h)^2 + k^2 = r^2 \implies (2h)^2 + k^2 = r^2 \implies 4h^2 + k^2 = r^2 \]

2. From point B:
\[ h^2 + (3k - k)^2 = r^2 \implies h^2 + (2k)^2 = r^2 \implies h^2 + 4k^2 = r^2 \]

### Step 7: Combine the Equations
Now we have two equations:
1. \( 4h^2 + k^2 = r^2 \)
2. \( h^2 + 4k^2 = r^2 \)

Setting them equal to each other gives:
\[ 4h^2 + k^2 = h^2 + 4k^2 \]

### Step 8: Solve for the Locus
Rearranging the equation:
\[ 4h^2 - h^2 = 4k^2 - k^2 \implies 3h^2 = 3k^2 \implies h^2 = k^2 \]

This implies:
\[ h = k \quad \text{or} \quad h = -k \]

### Step 9: Final Equation
Using \( h^2 + k^2 = r^2 \), we can express the locus of the centroid:
\[ x^2 + y^2 = \frac{r^2}{9} \]

### Conclusion
The locus of the centroid of triangle OAB is a circle centered at the origin with radius \( \frac{r}{3} \).

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