Circle C touches the line `l_1=4x-3y+k_1=0 , l_2=4x-3y+k_2=0` where `k_1 , k_2 in R` . If a line passing through the centre of circle intersect at `l_1` at `(-1,2)` and `l_2` at `(3,-6)` equation of circle is:

Options

A. `x^2+y^2-2x+4y-11=0`
B. `x^2+y^2+2x-4y-11=0`
C. `x^2+y^2-2x+6y-11=0`
D. `x^2+y^2-2x-4y+11=0`

Answer

To solve the problem, we need to find the equation of a circle that touches two parallel lines and has its center on a line that intersects these lines at given points.

### Step-by-step Solution:

1. **Identify the equations of the lines:**
The equations of the lines are given as:
\[ l_1: 4x - 3y + k_1 = 0 \]
\[ l_2: 4x - 3y + k_2 = 0 \]

2. **Find the values of \( k_1 \) and \( k_2 \):**
We know that the line \( l_1 \) intersects at the point \((-1, 2)\):
\[ 4(-1) - 3(2) + k_1 = 0 \]
Simplifying this:
\[ -4 - 6 + k_1 = 0 \implies k_1 = 10 \]
Now for \( l_2 \) at the point \((3, -6)\):
\[ 4(3) - 3(-6) + k_2 = 0 \]
Simplifying this:
\[ 12 + 18 + k_2 = 0 \implies k_2 = -30 \]

3. **Write the equations of the lines with the found constants:**
Thus, the equations of the lines become:
\[ l_1: 4x - 3y + 10 = 0 \]
\[ l_2: 4x - 3y - 30 = 0 \]

4. **Find the center of the circle:**
The center of the circle lies on the line that intersects \( l_1 \) and \( l_2 \) at the points \((-1, 2)\) and \((3, -6)\). We find the midpoint of these two points:
\[ C\left(\frac{-1 + 3}{2}, \frac{2 - 6}{2}\right) = C\left(1, -2\right) \]

5. **Calculate the distance between the two lines:**
The distance \( D \) between two parallel lines \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \) is given by:
\[ D = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
Here, \( A = 4, B = -3, C_1 = 10, C_2 = -30 \):
\[ D = \frac{|-30 - 10|}{\sqrt{4^2 + (-3)^2}} = \frac{40}{\sqrt{16 + 9}} = \frac{40}{5} = 8 \]

6. **Find the radius of the circle:**
Since the circle touches both lines, the radius \( r \) is half the distance between the lines:
\[ r = \frac{D}{2} = \frac{8}{2} = 4 \]

7. **Write the equation of the circle:**
The general equation of a circle with center \((h, k)\) and radius \( r \) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 1, k = -2, r = 4 \):
\[ (x - 1)^2 + (y + 2)^2 = 16 \]

8. **Expand the equation:**
Expanding this gives:
\[ (x^2 - 2x + 1) + (y^2 + 4y + 4) = 16 \]
Simplifying:
\[ x^2 + y^2 - 2x + 4y + 5 - 16 = 0 \implies x^2 + y^2 - 2x + 4y - 11 = 0 \]

### Final Equation of the Circle:
\[ x^2 + y^2 - 2x + 4y - 11 = 0 \]