Let a circle C touch the lines `L_1 : 4x-3y +K_1=0 and L_2: 4x-3y + K_2=0, K_1, K_2 in`R. If a line passing through the centre of the circle C intersects `L_1` at `(-1, 2)` and `L_2` at (3, -6), then the equation of the circle C is :

Options

A. `(x - 1)^2+(y-2)^2=4`
B. `(x + 1)^2+(y-2)^2=4`
C. `(x - 1)^2+(y+2)^2=16`
D. `(x - 1)^2+(y-2)^2=16`

Answer

To solve the problem, we need to find the equation of a circle that touches two parallel lines and passes through a specific point. Here's a step-by-step solution:

### Step 1: Identify the equations of the lines
The equations of the lines are given as:
- \( L_1: 4x - 3y + K_1 = 0 \)
- \( L_2: 4x - 3y + K_2 = 0 \)

### Step 2: Determine the values of \( K_1 \) and \( K_2 \)
We know that the line intersects \( L_1 \) at the point \( (-1, 2) \). Substituting these coordinates into the equation of \( L_1 \):
\[ 4(-1) - 3(2) + K_1 = 0 \]
\[ -4 - 6 + K_1 = 0 \implies K_1 = 10 \]

Next, we check the intersection with \( L_2 \) at the point \( (3, -6) \):
\[ 4(3) - 3(-6) + K_2 = 0 \]
\[ 12 + 18 + K_2 = 0 \implies K_2 = -30 \]

### Step 3: Write the equations of the lines with found constants
Now we can write the equations of the lines:
- \( L_1: 4x - 3y + 10 = 0 \)
- \( L_2: 4x - 3y - 30 = 0 \)

### Step 4: Find the center of the circle
The center of the circle lies on the line that passes through points \( (-1, 2) \) and \( (3, -6) \). To find the midpoint (which is the center of the circle), we use the midpoint formula:
\[ \text{Midpoint} = \left( \frac{-1 + 3}{2}, \frac{2 - 6}{2} \right) = \left( \frac{2}{2}, \frac{-4}{2} \right) = (1, -2) \]

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

### Step 6: Determine the radius of the circle
Since the circle touches both lines, the radius \( r \) of the circle is half the distance between the two lines:
\[ r = \frac{d}{2} = \frac{8}{2} = 4 \]

### Step 7: 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 \( (h, k) = (1, -2) \) and \( r = 4 \):
\[ (x - 1)^2 + (y + 2)^2 = 4^2 \]
\[ (x - 1)^2 + (y + 2)^2 = 16 \]

### Step 8: Expand the equation
Expanding the equation:
\[ (x^2 - 2x + 1) + (y^2 + 4y + 4) = 16 \]
Combining like terms:
\[ x^2 + y^2 - 2x + 4y + 5 - 16 = 0 \]
\[ x^2 + y^2 - 2x + 4y - 11 = 0 \]

Thus, the equation of the circle is:
\[ \boxed{x^2 + y^2 - 2x + 4y - 11 = 0} \]