The equation of the circle which passes through the origin and the points of intersection of the circle `x^(2)+y^(2)=4` and the line `x + y = 2` is

Options

A. `x^(2)+y^(2)= 4(x+y)`
B. `x^(2)+y^(2)=2(x+y)`
C. `x^(2)+y^(2) =3(x+y)`
D. `x^(2)+y^(2) =(x+y)`

Answer

To find the equation of the circle that passes through the origin and the points of intersection of the circle \(x^2 + y^2 = 4\) and the line \(x + y = 2\), we can follow these steps:

### Step 1: Identify the given equations
We have:
1. Circle: \(x^2 + y^2 = 4\) (which can be rewritten as \(x^2 + y^2 - 4 = 0\))
2. Line: \(x + y = 2\) (which can be rewritten as \(x + y - 2 = 0\))

### Step 2: Set up the general equation of the circle
The general equation of a circle that passes through the points of intersection of the given circle and line can be expressed as:
\[ S + \lambda L = 0 \]
Where \(S\) is the equation of the circle and \(L\) is the equation of the line. Thus, we have:
\[ x^2 + y^2 - 4 + \lambda (x + y - 2) = 0 \]
This simplifies to:
\[ x^2 + y^2 - 4 + \lambda x + \lambda y - 2\lambda = 0 \]

### Step 3: Substitute the origin into the equation
Since the circle passes through the origin (0, 0), we substitute \(x = 0\) and \(y = 0\) into the equation:
\[ 0^2 + 0^2 - 4 + \lambda(0) + \lambda(0) - 2\lambda = 0 \]
This simplifies to:
\[ -4 - 2\lambda = 0 \]

### Step 4: Solve for \(\lambda\)
Rearranging the equation gives:
\[ -2\lambda = 4 \implies \lambda = -2 \]

### Step 5: Substitute \(\lambda\) back into the circle equation
Now we substitute \(\lambda = -2\) back into the equation we derived in Step 2:
\[ x^2 + y^2 - 4 - 2(x + y - 2) = 0 \]
This expands to:
\[ x^2 + y^2 - 4 - 2x - 2y + 4 = 0 \]
Simplifying this gives:
\[ x^2 + y^2 - 2x - 2y = 0 \]

### Step 6: Final equation of the circle
The final equation of the circle is:
\[ x^2 + y^2 - 2x - 2y = 0 \]