If one end of a diameter of the circle `3x^(2)+3y^(2)-9x+6y+5=0` is (1,2), then the other end is

Options

A. (2,1)
B. (2,4)
C. (2,-4)
D. (-4,2)

Answer

To find the other end of the diameter of the circle given the equation \(3x^2 + 3y^2 - 9x + 6y + 5 = 0\) and one end of the diameter as \((1, 2)\), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we will rewrite the equation of the circle in standard form. We can factor out the 3 from the equation:

\[ 3(x^2 + y^2 - 3x + 2y) + 5 = 0 \]

This simplifies to:

\[ x^2 + y^2 - 3x + 2y + \frac{5}{3} = 0 \]

### Step 2: Complete the Square
Next, we will complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 - 3x \quad \text{can be written as} \quad (x - \frac{3}{2})^2 - \frac{9}{4} \]

For \(y\):
\[ y^2 + 2y \quad \text{can be written as} \quad (y + 1)^2 - 1 \]

Substituting these back into the equation gives:

\[ \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} + \left(y + 1\right)^2 - 1 + \frac{5}{3} = 0 \]

### Step 3: Combine and Simplify
Now we need to combine the constants:

\[ -\frac{9}{4} - 1 + \frac{5}{3} = -\frac{9}{4} - \frac{4}{4} + \frac{20}{12} = -\frac{13}{4} + \frac{20}{12} \]

Finding a common denominator (which is 12):

\[ -\frac{39}{12} + \frac{20}{12} = -\frac{19}{12} \]

Thus, the equation becomes:

\[ \left(x - \frac{3}{2}\right)^2 + \left(y + 1\right)^2 = \frac{19}{12} \]

### Step 4: Identify the Center
From the standard form of the circle \((x - h)^2 + (y - k)^2 = r^2\), we can identify the center \((h, k)\):

\[ h = \frac{3}{2}, \quad k = -1 \]

### Step 5: Use the Midpoint Formula
The center of the circle is the midpoint of the diameter. We know one endpoint of the diameter is \((1, 2)\). Let the other endpoint be \((x_2, y_2)\). The midpoint formula gives us:

\[ \left(\frac{1 + x_2}{2}, \frac{2 + y_2}{2}\right) = \left(\frac{3}{2}, -1\right) \]

### Step 6: Set Up the Equations
Setting up the equations from the midpoint:

1. \(\frac{1 + x_2}{2} = \frac{3}{2}\)
2. \(\frac{2 + y_2}{2} = -1\)

### Step 7: Solve for \(x_2\) and \(y_2\)
From the first equation:

\[ 1 + x_2 = 3 \implies x_2 = 2 \]

From the second equation:

\[ 2 + y_2 = -2 \implies y_2 = -4 \]

### Conclusion
Thus, the coordinates of the other end of the diameter are \((2, -4)\).