A circle passes through origin and meets the axes at A and B so that (2,3) lies on `bar(AB)` then the locus of centroid of `DeltaOAB` is

Options

A. `2x-3y=6xy`
B. `2x+3y=6xy`
C. `3x-2y=3xy`
D. `3x+2y=3xy`

Answer

To solve the problem step by step, we will find the locus of the centroid of triangle OAB, where O is the origin, A is on the x-axis, and B is on the y-axis.

### Step 1: Define Points A and B
Let the coordinates of point A be (A, 0) and the coordinates of point B be (0, B). The circle passes through the origin (0, 0) and meets the axes at these points.

### Step 2: Equation of Line AB
The line AB can be expressed in intercept form as:
\[ \frac{x}{A} + \frac{y}{B} = 1 \]

### Step 3: Substitute the Point (2, 3)
Since the point (2, 3) lies on the line AB, we substitute x = 2 and y = 3 into the line equation:
\[ \frac{2}{A} + \frac{3}{B} = 1 \]

### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[ 2B + 3A = AB \]

### Step 5: Finding the Centroid of Triangle OAB
The centroid (H, K) of triangle OAB is given by the average of the coordinates of the vertices:
\[ H = \frac{0 + A + 0}{3} = \frac{A}{3} \]
\[ K = \frac{0 + 0 + B}{3} = \frac{B}{3} \]

### Step 6: Express A and B in terms of H and K
From the equations for H and K, we can express A and B as:
\[ A = 3H \]
\[ B = 3K \]

### Step 7: Substitute A and B into the Line Equation
Substituting A and B back into the equation \(2B + 3A = AB\):
\[ 2(3K) + 3(3H) = (3H)(3K) \]
This simplifies to:
\[ 6K + 9H = 9HK \]

### Step 8: Rearranging the Equation
Rearranging gives us:
\[ 9HK - 9H - 6K = 0 \]
Dividing through by 3:
\[ 3HK - 3H - 2K = 0 \]

### Step 9: Final Equation
This can be rearranged to express the locus of the centroid:
\[ 3H + 2K = 3HK \]
Substituting H with x and K with y, we get:
\[ 3x + 2y = 3xy \]

### Final Result
Thus, the locus of the centroid of triangle OAB is given by the equation:
\[ 3x + 2y = 3xy \]