The equation of the circle which passes through the points of intersection of the circles `x^(2)+y^(2)-6x=0` and `x^(2)+y^(2)-6y=0` and has its centre at (3/2, 3/2), is

Options

A. `x^(2)+y^(2)+3x+3y+9=0`
B. `x^(2)+y^(2)+3x+3y=0`
C. `x^(2)+y^(2)-3x-3y=0`
D. `x^(2)+y^(2)-3x-3y+9=0`

Explanation

To find the equation of the circle that passes through the points of intersection of the circles given by the equations \( x^2 + y^2 - 6x = 0 \) and \( x^2 + y^2 - 6y = 0 \), and has its center at \( \left( \frac{3}{2}, \frac{3}{2} \right) \), we can follow these steps:

### Step 1: Find the equations of the given circles
The first circle can be rewritten as:
\[ x^2 + y^2 - 6x = 0 \implies (x - 3)^2 + y^2 = 9 \]
This represents a circle centered at \( (3, 0) \) with a radius of \( 3 \).

The second circle can be rewritten as:
\[ x^2 + y^2 - 6y = 0 \implies x^2 + (y - 3)^2 = 9 \]
This represents a circle centered at \( (0, 3) \) with a radius of \( 3 \).

### Step 2: Find the points of intersection of the two circles
To find the points of intersection, we can set the two equations equal to each other:
\[ x^2 + y^2 - 6x = x^2 + y^2 - 6y \]
This simplifies to:
\[ -6x = -6y \implies x = y \]
Now substitute \( y = x \) into one of the original circle equations:
\[ x^2 + x^2 - 6x = 0 \implies 2x^2 - 6x = 0 \implies 2x(x - 3) = 0 \]
Thus, \( x = 0 \) or \( x = 3 \). Therefore, the points of intersection are:
\[ (0, 0) \quad \text{and} \quad (3, 3) \]

### Step 3: Use the family of circles concept
The equation of the family of circles passing through the points of intersection can be expressed as:
\[ x^2 + y^2 - 6x + \lambda (x^2 + y^2 - 6y) = 0 \]
This simplifies to:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 - 6(1 - \lambda)y = 0 \]

### Step 4: Center of the circle
Since the center of the circle is given as \( \left( \frac{3}{2}, \frac{3}{2} \right) \), we can express the center in terms of \( \lambda \):
\[ \text{Center} = \left( -\frac{6(1 - \lambda)}{2(1 + \lambda)}, -\frac{6(1 - \lambda)}{2(1 + \lambda)} \right) \]
Setting this equal to \( \left( \frac{3}{2}, \frac{3}{2} \right) \):
\[ -\frac{6(1 - \lambda)}{2(1 + \lambda)} = \frac{3}{2} \]
Cross-multiplying gives:
\[ -6(1 - \lambda) = 3(1 + \lambda) \implies -6 + 6\lambda = 3 + 3\lambda \]
Rearranging gives:
\[ 3\lambda = 9 \implies \lambda = 3 \]

### Step 5: Substitute \( \lambda \) back into the equation
Substituting \( \lambda = 3 \) into the family of circles equation:
\[ x^2 + y^2 - 6x + 3(x^2 + y^2 - 6y) = 0 \]
This simplifies to:
\[ 4x^2 + 4y^2 - 6x - 18y = 0 \]
Dividing through by 2:
\[ 2x^2 + 2y^2 - 3x - 9y = 0 \]

### Step 6: Rearranging to standard form
Rearranging gives:
\[ 2x^2 + 2y^2 - 3x - 9y + 0 = 0 \]

### Final Equation
The equation of the circle is:
\[ 2x^2 + 2y^2 - 3x - 9y = 0 \]

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