The circle with centre (1,-3) and radius being the distance between the parallel lines `3x-4y-2=0, 3x-4y+8=0` is

Options

A. `x^(2)+y^(2)-2x+6y+6=0`
B. `x^(2)+y^(2)-2x+6y+8=0`
C. `x^(2)+y^(2)-2x+6y-90=0`
D. `x^(2)+y^(2)+2x-6y+6=0`

Answer

To solve the problem, we need to find the equation of the circle with a given center and radius, where the radius is determined by the distance between two parallel lines.

### Step-by-Step Solution:

1. **Identify the center of the circle:**
The center of the circle is given as \( (1, -3) \).

2. **Identify the equations of the parallel lines:**
The equations of the parallel lines are:
\[ 3x - 4y - 2 = 0 \quad \text{(Line 1)} \]
\[ 3x - 4y + 8 = 0 \quad \text{(Line 2)} \]

3. **Use the formula for the distance between two parallel lines:**
The formula for the distance \( d \) between two parallel lines of the form \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \) is:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
Here, \( A = 3 \), \( B = -4 \), \( C_1 = -2 \), and \( C_2 = 8 \).

4. **Calculate the distance:**
Substitute the values into the formula:
\[ d = \frac{|8 - (-2)|}{\sqrt{3^2 + (-4)^2}} = \frac{|8 + 2|}{\sqrt{9 + 16}} = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2 \]
Thus, the radius \( r \) of the circle is \( 2 \).

5. **Write the equation of the circle:**
The standard form of the equation of a circle with center \( (h, k) \) and radius \( r \) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 1 \), \( k = -3 \), and \( r = 2 \):
\[ (x - 1)^2 + (y + 3)^2 = 2^2 \]
This simplifies to:
\[ (x - 1)^2 + (y + 3)^2 = 4 \]

6. **Expand the equation:**
Expanding the left side:
\[ (x^2 - 2x + 1) + (y^2 + 6y + 9) = 4 \]
Combine like terms:
\[ x^2 + y^2 - 2x + 6y + 10 = 4 \]
Rearranging gives:
\[ x^2 + y^2 - 2x + 6y + 6 = 0 \]

### Final Equation:
The equation of the circle is:
\[ x^2 + y^2 - 2x + 6y + 6 = 0 \]