A circle of radius r passes through the origin and meets the axes at A and B. The locus of the centroid of `triangleOAB" is"`

Options

A. `x^(2)+y^(2)=4r^(2)`
B. `x^(2)+y^(2)=3r^(2)`
C. `3(x^(2)+y^(2))=r^(2)`
D. `9(x^(2)+y^(2))=4r^(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 meets the axes at points A and B, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circle**:
The circle passes through the origin (0, 0) and intersects the x-axis and y-axis at points A and B respectively. Let the coordinates of A be \( (a, 0) \) and the coordinates of B be \( (0, b) \).

2. **Equation of the Circle**:
Since the circle has a radius \( r \) and passes through the origin, its equation can be written as:
\[ x^2 + y^2 = r^2 \]
The points A and B must satisfy this equation.

3. **Coordinates of Points A and B**:
From the circle's equation, we know:
- For point A on the x-axis: \( a^2 + 0^2 = r^2 \) implies \( a^2 = r^2 \) so \( a = r \) or \( a = -r \).
- For point B on the y-axis: \( 0^2 + b^2 = r^2 \) implies \( b^2 = r^2 \) so \( b = r \) or \( b = -r \).

4. **Centroid of Triangle OAB**:
The centroid \( G \) of triangle OAB is given by the formula:
\[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Here, the vertices are \( O(0, 0) \), \( A(a, 0) \), and \( B(0, b) \). Thus, the coordinates of the centroid are:
\[ G\left( \frac{0 + a + 0}{3}, \frac{0 + 0 + b}{3} \right) = \left( \frac{a}{3}, \frac{b}{3} \right) \]

5. **Expressing a and b in terms of the centroid**:
Let \( G \) be \( (H, K) \). Then:
\[ H = \frac{a}{3} \quad \text{and} \quad K = \frac{b}{3} \]
This implies:
\[ a = 3H \quad \text{and} \quad b = 3K \]

6. **Using the Circle Equation**:
Since \( A \) and \( B \) lie on the circle, we can substitute \( a \) and \( b \) into the circle's equation:
\[ a^2 + b^2 = r^2 \]
Substituting \( a = 3H \) and \( b = 3K \):
\[ (3H)^2 + (3K)^2 = r^2 \]
Simplifying this gives:
\[ 9H^2 + 9K^2 = r^2 \]

7. **Final Locus Equation**:
Rearranging gives:
\[ H^2 + K^2 = \frac{r^2}{9} \]
Replacing \( H \) with \( x \) and \( K \) with \( y \):
\[ 9x^2 + y^2 = r^2 \]

### Conclusion:
The locus of the centroid of triangle OAB is given by the equation:
\[ 9x^2 + y^2 = r^2 \]