Locus of the centre of the circle touching the line `3x + 4y + 1=0` and having radius equal to 5 units is

Options

A. a pair of perpendicular lines
B. a pair of parallel lines
C. a pair of straight lines
D. none of these

Answer

To find the locus of the center of a circle that touches the line \(3x + 4y + 1 = 0\) and has a radius of 5 units, we can follow these steps:

### Step-by-step Solution:

1. **Identify the Line and Circle Parameters**:
- The equation of the line is given as \(3x + 4y + 1 = 0\).
- The radius of the circle \(r = 5\) units.

2. **Determine the Perpendicular Distance Formula**:
- The formula for the perpendicular distance \(d\) from a point \((h, k)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
- Here, \(A = 3\), \(B = 4\), and \(C = 1\).

3. **Substitute the Center of the Circle**:
- The center of the circle is denoted as \((h, k)\).
- The distance from the center to the line must equal the radius, so we set up the equation:
\[ \frac{|3h + 4k + 1|}{\sqrt{3^2 + 4^2}} = 5 \]
- Calculate \(\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\).

4. **Set Up the Equation**:
- The equation simplifies to:
\[ |3h + 4k + 1| = 5 \cdot 5 = 25 \]
- This leads to two cases:
\[ 3h + 4k + 1 = 25 \quad \text{(1)} \]
\[ 3h + 4k + 1 = -25 \quad \text{(2)} \]

5. **Solve Each Case**:
- For case (1):
\[ 3h + 4k = 24 \quad \text{(Equation 1)} \]
- For case (2):
\[ 3h + 4k = -26 \quad \text{(Equation 2)} \]

6. **Identify the Locus**:
- The equations \(3h + 4k = 24\) and \(3h + 4k = -26\) represent two parallel lines in the \(hk\)-plane (where \(h\) and \(k\) are the coordinates of the center of the circle).
- Therefore, the locus of the center of the circle is a pair of parallel lines.

### Final Answer:
The locus of the center of the circle is a pair of parallel lines given by the equations:
\[ 3x + 4y = 24 \quad \text{and} \quad 3x + 4y = -26 \]