Let the lines `4x-3y+10=0` and `4x-3y-30=0` make equal intercepts of 6 units with a circle (C ) whose centre lies on `2x+y=0`, then the equation of the circle C is

Options

A. `x^(2)+y^(2)-2x+4y-20=0`
B. `x^(2)+y^(2)-4x+8y-20=0`
C. `x^(2)+y^(2)+2x-4y-20=0`
D. `x^(2)+y^(2)-4x+8y-5=0`

Answer

To solve the problem step by step, we will follow the given information and derive the equation of the circle \( C \).

### Step 1: Identify the lines and their properties
We have two lines given by the equations:
1. \( 4x - 3y + 10 = 0 \)
2. \( 4x - 3y - 30 = 0 \)

These lines are parallel because they have the same coefficients for \( x \) and \( y \).

### Step 2: Find the distance between the two parallel lines
The distance \( d \) between two parallel lines \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \) is given by the formula:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
For our lines:
- \( C_1 = 10 \)
- \( C_2 = -30 \)
- \( A = 4 \)
- \( B = -3 \)

Calculating the distance:
\[ d = \frac{|(-30) - 10|}{\sqrt{4^2 + (-3)^2}} = \frac{|-40|}{\sqrt{16 + 9}} = \frac{40}{5} = 8 \]

### Step 3: Determine the center of the circle
The center of the circle lies on the line \( 2x + y = 0 \). We can express \( y \) in terms of \( x \):
\[ y = -2x \]

Let the center of the circle be \( (a, -2a) \).

### Step 4: Find the intercepts of the lines with the circle
The lines make equal intercepts of 6 units with the circle. This means the radius of the circle must be equal to half the distance between the two lines, which is \( 4 \) units (since \( \frac{8}{2} = 4 \)).

### Step 5: Use the Pythagorean theorem to find the radius
The radius \( r \) of the circle can be found using the distance from the center to one of the lines. The radius is also the hypotenuse of a right triangle formed by the distance from the center to the line and the intercepts.

Using the distance formula from a point to a line, the distance \( D \) from the point \( (a, -2a) \) to the line \( 4x - 3y + 10 = 0 \) is given by:
\[ D = \frac{|4a - 3(-2a) + 10|}{\sqrt{4^2 + (-3)^2}} = \frac{|4a + 6a + 10|}{5} = \frac{|10a + 10|}{5} = 2|a + 1| \]

Setting this equal to the radius \( 4 \):
\[ 2|a + 1| = 4 \implies |a + 1| = 2 \]

This gives us two cases:
1. \( a + 1 = 2 \) → \( a = 1 \)
2. \( a + 1 = -2 \) → \( a = -3 \)

### Step 6: Find the corresponding \( y \) values
For \( a = 1 \):
\[ y = -2(1) = -2 \quad \Rightarrow \quad \text{Center} = (1, -2) \]

For \( a = -3 \):
\[ y = -2(-3) = 6 \quad \Rightarrow \quad \text{Center} = (-3, 6) \]

### Step 7: Write the equations of the circles
The radius \( r = 4 \). The equations of the circles are:
1. For center \( (1, -2) \):
\[ (x - 1)^2 + (y + 2)^2 = 16 \]
2. For center \( (-3, 6) \):
\[ (x + 3)^2 + (y - 6)^2 = 16 \]

### Final Answer
The equation of the circle \( C \) can be expressed as:
1. \( (x - 1)^2 + (y + 2)^2 = 16 \)
2. \( (x + 3)^2 + (y - 6)^2 = 16 \)