Which of the following line is a diameter of the circel `x ^(2) + y^(2) -6x-8y-9=0` ?

Options

A. `3x-4y =0`
B. `4x-y=0`
C. `x+y=7`
D. `x-y=1`

Answer

To determine which of the given lines is a diameter of the circle defined by the equation \(x^2 + y^2 - 6x - 8y - 9 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation in Standard Form
We start with the equation of the circle:
\[ x^2 + y^2 - 6x - 8y - 9 = 0 \]
To convert it into standard form, we complete the square for both \(x\) and \(y\).

1. Rearranging the terms:
\[ x^2 - 6x + y^2 - 8y = 9 \]

2. Completing the square for \(x\):
- Take half of \(-6\) (which is \(-3\)), square it (which is \(9\)), and add it inside the equation:
\[ (x - 3)^2 - 9 \]

3. Completing the square for \(y\):
- Take half of \(-8\) (which is \(-4\)), square it (which is \(16\)), and add it inside the equation:
\[ (y - 4)^2 - 16 \]

4. Substitute back into the equation:
\[ (x - 3)^2 - 9 + (y - 4)^2 - 16 = 9 \]
Simplifying gives:
\[ (x - 3)^2 + (y - 4)^2 = 34 \]

### Step 2: Identify the Center and Radius of the Circle
From the standard form \((x - 3)^2 + (y - 4)^2 = 34\), we can identify:
- The center of the circle is \((3, 4)\).
- The radius is \(\sqrt{34}\).

### Step 3: Check Each Line Option
We need to check which of the given lines passes through the center \((3, 4)\). The lines given are:

1. **Option A:** \(3x - 4y = 0\)
- Substitute \(x = 3\) and \(y = 4\):
\[ 3(3) - 4(4) = 9 - 16 = -7 \quad (\text{not } 0) \]
So, Option A is incorrect.

2. **Option B:** \(4x - y = 0\)
- Substitute \(x = 3\) and \(y = 4\):
\[ 4(3) - 4 = 12 - 4 = 8 \quad (\text{not } 0) \]
So, Option B is incorrect.

3. **Option C:** \(x + y = 7\)
- Substitute \(x = 3\) and \(y = 4\):
\[ 3 + 4 = 7 \quad (\text{equals } 7) \]
So, Option C is correct.

4. **Option D:** \(x - y = 1\)
- Substitute \(x = 3\) and \(y = 4\):
\[ 3 - 4 = -1 \quad (\text{not } 1) \]
So, Option D is incorrect.

### Conclusion
The line that is a diameter of the circle is **Option C: \(x + y = 7\)**.

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