The circles `x^2+y^2+x+y=0` and `x^2+y^2+x-y=0` intersect at an angle of

Options

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

Answer

To find the angle at which the circles \(x^2 + y^2 + x + y = 0\) and \(x^2 + y^2 + x - y = 0\) intersect, we will follow these steps:

### Step 1: Rewrite the equations of the circles
The given equations can be rewritten in standard form.

1. For the first circle:
\[ x^2 + y^2 + x + y = 0 \implies x^2 + y^2 + x + y + \frac{1}{2} = \frac{1}{2} \implies \left(x + \frac{1}{2}\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{1}{2} \]
This represents a circle centered at \((-0.5, -0.5)\) with radius \(\frac{1}{\sqrt{2}}\).

2. For the second circle:
\[ x^2 + y^2 + x - y = 0 \implies x^2 + y^2 + x + \frac{1}{2} = \frac{1}{2} \implies \left(x + \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{2} \]
This represents a circle centered at \((-0.5, 0.5)\) with radius \(\frac{1}{\sqrt{2}}\).

### Step 2: Find the points of intersection
To find the points of intersection, we set the equations equal to each other:
\[ x^2 + y^2 + x + y = x^2 + y^2 + x - y \]
This simplifies to:
\[ y = 0 \]
Now, substituting \(y = 0\) into either circle's equation:
\[ x^2 + 0^2 + x + 0 = 0 \implies x^2 + x = 0 \implies x(x + 1) = 0 \]
Thus, the points of intersection are:
\[ (0, 0) \quad \text{and} \quad (-1, 0) \]

### Step 3: Find the tangents at the points of intersection
For the first circle at point \((0, 0)\):
Using the formula for the tangent line:
\[ T = 0 \implies -1 \cdot x + 0 \cdot y + 0 + 0 = 0 \implies x = 0 \]
The tangent line is vertical.

For the second circle at point \((0, 0)\):
Using the formula for the tangent line:
\[ T = 0 \implies -1 \cdot x + 0 \cdot y + 0 - 0 = 0 \implies x = 0 \]
The tangent line is also vertical.

### Step 4: Find the slopes of the tangents
The slopes of the tangents at the points of intersection are:
- For the first circle: \(m_1 = \infty\) (vertical line)
- For the second circle: \(m_2 = \infty\) (vertical line)

### Step 5: Calculate the angle of intersection
The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan(\theta) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]
Since both slopes are infinite, we can conclude that the angle of intersection is:
\[ \theta = 90^\circ \quad \text{or} \quad \frac{\pi}{2} \text{ radians} \]

### Final Answer
The circles intersect at an angle of \(\frac{\pi}{2}\) radians.

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