A variable line cuts X-axis at A, Y -axix at B, where OA = a, OB = b (O as origin) such that `a^(2)+b^(2)=1`.
Find the locus of
centroid of `Delta OAB`

Answer

To find the locus of the centroid of triangle OAB, where the line cuts the X-axis at point A and the Y-axis at point B, we can follow these steps:

### Step 1: Identify the coordinates of points A and B
- The point A, where the line intersects the X-axis, has coordinates (a, 0).
- The point B, where the line intersects the Y-axis, has coordinates (0, b).
- The origin O has coordinates (0, 0).

### Step 2: Write the formula for the centroid of triangle OAB
The centroid (G) of a triangle with vertices at (x1, y1), (x2, y2), and (x3, y3) is given by:
\[ G = \left( \frac{x1 + x2 + x3}{3}, \frac{y1 + y2 + y3}{3} \right) \]
For triangle OAB, the coordinates are:
- O(0, 0)
- A(a, 0)
- B(0, b)

Thus, the coordinates of the centroid G are:
\[ G = \left( \frac{0 + a + 0}{3}, \frac{0 + 0 + b}{3} \right) = \left( \frac{a}{3}, \frac{b}{3} \right) \]

### Step 3: Express a and b in terms of the centroid coordinates
Let the coordinates of the centroid be (x, y). Then we have:
\[ x = \frac{a}{3} \quad \text{and} \quad y = \frac{b}{3} \]
From these equations, we can express a and b as:
\[ a = 3x \quad \text{and} \quad b = 3y \]

### Step 4: Use the given condition \(a^2 + b^2 = 1\)
According to the problem, we have the condition:
\[ a^2 + b^2 = 1 \]
Substituting the expressions for a and b:
\[ (3x)^2 + (3y)^2 = 1 \]
This simplifies to:
\[ 9x^2 + 9y^2 = 1 \]

### Step 5: Simplify the equation
Dividing the entire equation by 9 gives:
\[ x^2 + y^2 = \frac{1}{9} \]

### Conclusion
The locus of the centroid G of triangle OAB is given by the equation:
\[ x^2 + y^2 = \frac{1}{9} \]
This represents a circle centered at the origin with a radius of \(\frac{1}{3}\).

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