The angle of intersection of the circles `x^(2)+y^(2)=4` and `x^(2)+y^(2)+2x+2y`, is

Options

A. `pi//2`
B. `pi//3`
C. `pi//6`
D. `pi//4`

Answer

To find the angle of intersection of the circles given by the equations \( x^2 + y^2 = 4 \) and \( x^2 + y^2 + 2x + 2y = 0 \), we will follow these steps:

### Step 1: Identify the equations of the circles
The first circle is given by:
\[ S_1: x^2 + y^2 = 4 \]
This can be rewritten in the standard form:
\[ x^2 + y^2 + 0x + 0y - 4 = 0 \]
From this, we identify:
- \( g_1 = 0 \)
- \( f_1 = 0 \)
- \( c_1 = -4 \)

The second circle is given by:
\[ S_2: x^2 + y^2 + 2x + 2y = 0 \]
This can be rewritten as:
\[ x^2 + y^2 + 2x + 2y + 0 = 0 \]
From this, we identify:
- \( g_2 = 1 \)
- \( f_2 = 1 \)
- \( c_2 = 0 \)

### Step 2: Use the formula for the angle of intersection
The formula for the cosine of the angle \( \theta \) of intersection of two circles is given by:
\[ \cos \theta = \frac{2g_1g_2 + 2f_1f_2 - c_1 - c_2}{2 \sqrt{(g_1^2 + f_1^2 - c_1)(g_2^2 + f_2^2 - c_2)}} \]

### Step 3: Substitute the values into the formula
Substituting the values we found:
- \( g_1 = 0 \), \( f_1 = 0 \), \( c_1 = -4 \)
- \( g_2 = 1 \), \( f_2 = 1 \), \( c_2 = 0 \)

We get:
\[ \cos \theta = \frac{2(0)(1) + 2(0)(1) - (-4) - 0}{2 \sqrt{(0^2 + 0^2 + 4)(1^2 + 1^2 - 0)}} \]
This simplifies to:
\[ \cos \theta = \frac{0 + 0 + 4}{2 \sqrt{(0 + 0 + 4)(1 + 1 - 0)}} \]
\[ = \frac{4}{2 \sqrt{4 \cdot 2}} \]
\[ = \frac{4}{2 \sqrt{8}} = \frac{4}{2 \cdot 2\sqrt{2}} = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 4: Find the angle
Since \( \cos \theta = \frac{1}{\sqrt{2}} \), we know that:
\[ \theta = \frac{\pi}{4} \]

### Conclusion
Thus, the angle of intersection of the circles is:
\[ \theta = \frac{\pi}{4} \]