Let `A (0,2 ) , B ( 3, 0 ) , C ( 6, 4 ) ` be the vertices of triangle and P is a point inside the triangle such that area of triangle APB, BPC and CPA are equal. Equation of circle circumscribing ` Delta APB ` is :

Options

A. ` x ^ 2 + y^ 2 - 2x - 3y = 0`
B. ` x ^ 2 + y ^ 2 - 4x - 3y = 0 `
C. ` x ^ 2 + y ^ 2 - 3x - 4y = 0`
D. ` x ^ 2 + y^ 2 - 3x - 2y = 0`

Explanation

To find the equation of the circle circumscribing triangle \( \Delta APB \) where \( A(0, 2) \), \( B(3, 0) \), and \( P \) is the centroid of triangle \( ABC \) with equal areas for triangles \( APB \), \( BPC \), and \( CPA \), we can follow these steps:

### Step 1: Find the coordinates of point P (the centroid)
The centroid \( P \) of triangle \( ABC \) can be calculated using the formula:
\[ P\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
where \( A(0, 2) \), \( B(3, 0) \), and \( C(6, 4) \).

Substituting the coordinates:
\[ P\left(\frac{0 + 3 + 6}{3}, \frac{2 + 0 + 4}{3}\right) = P\left(\frac{9}{3}, \frac{6}{3}\right) = P(3, 2) \]

### Step 2: Determine the equation of the circle
Since \( PA \) is perpendicular to \( PB \) and \( AB \) is the diameter of the circle, we can use the midpoint of \( AB \) to find the center of the circle.

1. **Find the midpoint of \( AB \)**:
\[ M\left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) = M\left(\frac{0 + 3}{2}, \frac{2 + 0}{2}\right) = M\left(\frac{3}{2}, 1\right) \]

2. **Find the radius**:
The radius \( r \) is the distance from the midpoint \( M \) to either point \( A \) or \( B \):
\[ r = \sqrt{\left(\frac{3}{2} - 0\right)^2 + \left(1 - 2\right)^2} = \sqrt{\left(\frac{3}{2}\right)^2 + (-1)^2} = \sqrt{\frac{9}{4} + 1} = \sqrt{\frac{9}{4} + \frac{4}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \]

3. **Equation of the circle**:
The standard equation of a circle with center \( (h, k) \) and radius \( r \) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here, \( h = \frac{3}{2} \), \( k = 1 \), and \( r^2 = \left(\frac{\sqrt{13}}{2}\right)^2 = \frac{13}{4} \).

Thus, the equation becomes:
\[ \left(x - \frac{3}{2}\right)^2 + (y - 1)^2 = \frac{13}{4} \]

Expanding this:
\[ \left(x^2 - 3x + \frac{9}{4}\right) + \left(y^2 - 2y + 1\right) = \frac{13}{4} \]
\[ x^2 + y^2 - 3x - 2y + \frac{9}{4} + 1 - \frac{13}{4} = 0 \]
\[ x^2 + y^2 - 3x - 2y - \frac{3}{4} = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 4x^2 + 4y^2 - 12x - 8y - 3 = 0 \]

### Final Equation
The final equation of the circle is:
\[ 4x^2 + 4y^2 - 12x - 8y - 3 = 0 \]