Find the equation of the circle passing through the intersection of the circles `x^2 + y^2-4 = 0` and `x^2+y^2-2x-4y+4=0` and touching the line `x + 2y=0`

Options

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

Answer

To find the equation of the circle passing through the intersection of the given circles and touching the specified line, we will follow these steps:

### Step 1: Identify the equations of the circles and the line
The equations of the circles are:
1. \( S_1: x^2 + y^2 - 4 = 0 \)
2. \( S_2: x^2 + y^2 - 2x - 4y + 4 = 0 \)

The equation of the line is:
\[ x + 2y = 0 \]

### Step 2: Write the general equation of the circle passing through the intersection of the two circles
The equation of the circle that passes through the intersection of the two given circles can be expressed as:
\[ S_1 + \lambda S_2 = 0 \]
where \( \lambda \) is a parameter.

### Step 3: Substitute the equations of the circles into the general equation
Substituting \( S_1 \) and \( S_2 \):
\[ (x^2 + y^2 - 4) + \lambda (x^2 + y^2 - 2x - 4y + 4) = 0 \]
This simplifies to:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 - 2\lambda x - 4\lambda y + (4\lambda - 4) = 0 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 - 2\lambda x - 4\lambda y + (4\lambda - 4) = 0 \]

### Step 5: Identify the center and radius of the circle
The center \( C \) of the circle can be found from the coefficients of \( x \) and \( y \):
- The center coordinates are:
\[ C_x = \frac{2\lambda}{1 + \lambda}, \quad C_y = \frac{4\lambda}{1 + \lambda} \]

### Step 6: Calculate the radius of the circle
The radius \( R \) can be expressed as:
\[ R = \sqrt{C_x^2 + C_y^2 - (4\lambda - 4)} \]
Substituting \( C_x \) and \( C_y \):
\[ R = \sqrt{\left(\frac{2\lambda}{1 + \lambda}\right)^2 + \left(\frac{4\lambda}{1 + \lambda}\right)^2 - (4\lambda - 4)} \]

### Step 7: Find the distance from the center to the line
The distance \( D \) from the center \( C \) to the line \( x + 2y = 0 \) is given by:
\[ D = \frac{|C_x + 2C_y|}{\sqrt{1^2 + 2^2}} = \frac{\left| \frac{2\lambda}{1 + \lambda} + 2 \cdot \frac{4\lambda}{1 + \lambda} \right|}{\sqrt{5}} = \frac{\left| \frac{10\lambda}{1 + \lambda} \right|}{\sqrt{5}} \]

### Step 8: Set the radius equal to the distance
Since the circle touches the line, we equate the radius \( R \) and the distance \( D \):
\[ R = D \]

### Step 9: Solve for \( \lambda \)
After equating and simplifying, we find the possible values for \( \lambda \). We find that \( \lambda = 1 \) is the valid solution.

### Step 10: Substitute \( \lambda \) back into the circle equation
Substituting \( \lambda = 1 \) back into the equation gives:
\[ x^2 + y^2 - 2x - 4y = 0 \]

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

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