If the equation of a circle of a circle is `10x^(2)-10x+y^(2)+6y=-9`, which of the following lines contains a diameter of the circle?

Options

A. `y=2x-7`
B. `y=-2x+7`
C. `y=2x+13`
D. `y=-2x+13`

Answer

To solve the problem, we need to rewrite the given equation of the circle in the standard form and then identify which line contains a diameter of the circle.

### Step 1: Rewrite the Circle's Equation
The given equation of the circle is:
\[ 10x^2 - 10x + y^2 + 6y = -9 \]

First, we can simplify this equation by dividing everything by 10:
\[ x^2 - x + \frac{y^2}{10} + \frac{3y}{5} = -\frac{9}{10} \]

Now, we will rearrange the equation:
\[ x^2 - x + y^2 + 6y + 9 = 0 \]

### Step 2: Complete the Square for x and y
Next, we will complete the square for the x and y terms.

For \(x\):
\[ x^2 - 10x \]
To complete the square:
\[ x^2 - 10x = (x - 5)^2 - 25 \]

For \(y\):
\[ y^2 + 6y \]
To complete the square:
\[ y^2 + 6y = (y + 3)^2 - 9 \]

### Step 3: Substitute Back into the Equation
Now, substituting back into the equation:
\[ (x - 5)^2 - 25 + (y + 3)^2 - 9 = -9 \]

This simplifies to:
\[ (x - 5)^2 + (y + 3)^2 - 34 = -9 \]
\[ (x - 5)^2 + (y + 3)^2 = 25 \]

### Step 4: Identify Center and Radius
From the equation \((x - 5)^2 + (y + 3)^2 = 25\), we can see that:
- The center of the circle is \((5, -3)\)
- The radius \(r = \sqrt{25} = 5\)

### Step 5: Determine the Diameter
A diameter of the circle will pass through the center. Therefore, we need to check which of the given lines passes through the center \((5, -3)\).

### Step 6: Check Each Option
Let's check each option to see which one contains the point \((5, -3)\).

1. **Option A**: Check if it contains (5, -3).
2. **Option B**: Check if it contains (5, -3).
3. **Option C**: Check if it contains (5, -3).
4. **Option D**: Check if it contains (5, -3).

After checking each option, we find that **Option B** satisfies the condition.

### Final Answer
The line that contains a diameter of the circle is **Option B**.