The equation of the diameter of the circle `3(x^(2)+y^(2))-2x+6y-9=0` which is perpendicular to the line 2x+3y=12 is

Options

A. 3x-2y+3=0
B. 3x-2y-3=0
C. 3x-2y+1=0
D. 3x-2y-1=0

Answer

To solve the problem, we need to find the equation of the diameter of the circle given by the equation \(3(x^2 + y^2) - 2x + 6y - 9 = 0\) that is perpendicular to the line \(2x + 3y = 12\).

### Step 1: Rewrite the Circle Equation
First, we simplify the equation of the circle by dividing everything by 3:

\[ x^2 + y^2 - \frac{2}{3}x + 2y - 3 = 0 \]

### Step 2: Identify the Center of the Circle
The standard form of a circle's equation is:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

From our equation, we can identify:

- \(2g = -\frac{2}{3} \Rightarrow g = -\frac{1}{3}\)
- \(2f = 2 \Rightarrow f = 1\)
- \(c = -3\)

Thus, the center of the circle \((h, k)\) is:

\[ \left(-g, -f\right) = \left(\frac{1}{3}, -1\right) \]

### Step 3: Find the Slope of the Given Line
The line given is \(2x + 3y = 12\). We can rewrite it in slope-intercept form \(y = mx + b\):

\[ 3y = -2x + 12 \Rightarrow y = -\frac{2}{3}x + 4 \]

The slope \(m\) of this line is \(-\frac{2}{3}\).

### Step 4: Find the Slope of the Diameter
Since the diameter is perpendicular to the line, the slope of the diameter \(m_d\) can be found using the negative reciprocal of the slope of the line:

\[ m_d = \frac{3}{2} \]

### Step 5: Write the Equation of the Diameter
Using the point-slope form of the equation of a line, we can write the equation of the diameter passing through the center \(\left(\frac{1}{3}, -1\right)\) with slope \(\frac{3}{2}\):

\[ y - y_1 = m(x - x_1) \]

Substituting \(y_1 = -1\), \(m = \frac{3}{2}\), and \(x_1 = \frac{1}{3}\):

\[ y + 1 = \frac{3}{2}\left(x - \frac{1}{3}\right) \]

### Step 6: Simplify the Equation
Now, we simplify this equation:

\[ y + 1 = \frac{3}{2}x - \frac{3}{6} \]
\[ y + 1 = \frac{3}{2}x - \frac{1}{2} \]
\[ y = \frac{3}{2}x - \frac{1}{2} - 1 \]
\[ y = \frac{3}{2}x - \frac{3}{2} \]

### Step 7: Convert to Standard Form
To convert this into standard form \(Ax + By + C = 0\):

\[ \frac{3}{2}x - y - \frac{3}{2} = 0 \]

Multiplying through by 2 to eliminate the fraction:

\[ 3x - 2y - 3 = 0 \]

### Final Answer
Thus, the equation of the diameter of the circle that is perpendicular to the line \(2x + 3y = 12\) is:

\[ 3x - 2y - 3 = 0 \]