For the circle `x^(2)+y^(2)-2x-4y-4=0`, then lines 2x+3y-1=0, 2x+y+5=0 are

Options

A. perpendicular tangents
B. conjugate
C. parallel tangents
D. perpendicular chords

Answer

To solve the problem, we need to analyze the given circle and the two lines to determine their relationship.

### Step-by-Step Solution:

1. **Rewrite the Circle Equation**:
The given equation of the circle is:
\[ x^2 + y^2 - 2x - 4y - 4 = 0 \]
We can rewrite it in the standard form \((x - a)^2 + (y - b)^2 = r^2\).

2. **Complete the Square**:
- For \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
- For \(y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]
- Substitute back into the circle equation:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 - 4 = 0 \]
\[ (x - 1)^2 + (y - 2)^2 - 9 = 0 \]
\[ (x - 1)^2 + (y - 2)^2 = 9 \]
This shows that the center of the circle is \((1, 2)\) and the radius \(r = 3\).

3. **Identify the Lines**:
The equations of the lines are:
- Line 1: \(2x + 3y - 1 = 0\)
- Line 2: \(2x + y + 5 = 0\)

4. **Convert the Lines to Standard Form**:
- Line 1 can be rewritten as:
\[ 3y = -2x + 1 \implies y = -\frac{2}{3}x + \frac{1}{3} \]
- Line 2 can be rewritten as:
\[ y = -2x - 5 \]

5. **Calculate Slopes**:
- Slope of Line 1 (\(m_1\)) = \(-\frac{2}{3}\)
- Slope of Line 2 (\(m_2\)) = \(-2\)

6. **Check for Perpendicularity**:
Two lines are perpendicular if the product of their slopes is \(-1\):
\[ m_1 \cdot m_2 = -\frac{2}{3} \cdot (-2) = \frac{4}{3} \neq -1 \]
Thus, the lines are not perpendicular.

7. **Check for Parallelism**:
Two lines are parallel if their slopes are equal:
\[ m_1 \neq m_2 \quad \text{(since } -\frac{2}{3} \neq -2\text{)} \]
Thus, the lines are not parallel.

8. **Check for Conjugate Lines**:
For the lines to be conjugate with respect to the circle, we use the condition:
\[ r^2 (l_1 l_2 + m_1 m_2) = n_1 n_2 \]
Here, \(r^2 = 9\), \(l_1 = 2\), \(m_1 = 3\), \(n_1 = 1\), \(l_2 = 2\), \(m_2 = 1\), and \(n_2 = 5\).

Calculate:
\[ l_1 l_2 + m_1 m_2 = 2 \cdot 2 + 3 \cdot 1 = 4 + 3 = 7 \]
\[ n_1 n_2 = 1 \cdot 5 = 5 \]
Now check:
\[ 9 \cdot 7 = 63 \quad \text{and} \quad 5 \neq 63 \]
Thus, they are not conjugate.

9. **Conclusion**:
Since the lines are neither perpendicular nor parallel, and they do not satisfy the conjugate condition, we conclude that the lines are **not conjugate**.

### Final Answer:
The lines \(2x + 3y - 1 = 0\) and \(2x + y + 5 = 0\) are neither perpendicular nor parallel nor conjugate to the given circle.

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