The equation of the circle which passes through the intersection of circle `x^(2)+y^(2)+4(x+y)+4=0` with the line `x+y+2=0` and has its centre at the origin is ............

Answer

To find the equation of the circle that passes through the intersection of the given circle and line, and has its center at the origin, we can follow these steps:

### Step 1: Write down the equations
The equation of the circle is given by:
\[ x^2 + y^2 + 4x + 4y + 4 = 0 \]
The equation of the line is:
\[ x + y + 2 = 0 \]

### Step 2: Rewrite the equations
We can rewrite the equation of the line as:
\[ y = -x - 2 \]

### Step 3: Find the intersection of the circle and the line
To find the intersection points, we substitute \( y = -x - 2 \) into the circle's equation:
\[ x^2 + (-x - 2)^2 + 4x + 4(-x - 2) + 4 = 0 \]
Expanding this:
\[ x^2 + (x^2 + 4x + 4) + 4x - 4x - 8 + 4 = 0 \]
Combine like terms:
\[ 2x^2 + 0x + 0 = 0 \]
This simplifies to:
\[ 2x^2 = 0 \]
Thus, \( x = 0 \). Substituting \( x = 0 \) back into the line equation gives:
\[ y = -0 - 2 = -2 \]
So, the intersection point is \( (0, -2) \).

### Step 4: General form of the circle
The general equation of a circle with center at the origin is:
\[ x^2 + y^2 = r^2 \]
We need to find \( r^2 \) such that the circle passes through the point \( (0, -2) \).

### Step 5: Substitute the intersection point into the circle equation
Substituting \( (0, -2) \) into the circle's equation:
\[ 0^2 + (-2)^2 = r^2 \]
This gives:
\[ 4 = r^2 \]

### Step 6: Write the final equation of the circle
Thus, the equation of the circle is:
\[ x^2 + y^2 = 4 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 = 4 \]