Angle of intersection of the following curves
`x^2+y^2=a^2sqrt2,x^2+y^2=a^2`, is `pi//4`.

Answer

To find the angle of intersection of the curves given by the equations \(x^2 + y^2 = a^2\sqrt{2}\) and \(x^2 + y^2 = a^2\), we will follow these steps:

### Step 1: Find the points of intersection of the curves
The curves are:
1. \(x^2 + y^2 = a^2\sqrt{2}\) (Curve 1)
2. \(x^2 + y^2 = a^2\) (Curve 2)

To find the points of intersection, we set the two equations equal to each other:
\[ a^2\sqrt{2} = a^2 \]
This simplifies to:
\[ \sqrt{2} = 1 \]
This is not possible, which means we need to find the points of intersection by substituting one equation into another.

### Step 2: Solve for \(y\) in terms of \(x\)
From the second equation:
\[ y^2 = a^2 - x^2 \]
Substituting this into the first equation:
\[ x^2 + (a^2 - x^2) = a^2\sqrt{2} \]
This simplifies to:
\[ a^2 = a^2\sqrt{2} \]
Again, this leads to a contradiction, indicating that we need to analyze the curves further.

### Step 3: Find the slopes of the tangents at the points of intersection
To find the slopes of the tangents, we will differentiate both equations implicitly.

For Curve 1:
\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(a^2\sqrt{2}) \]
This gives:
\[ 2x + 2y \frac{dy}{dx} = 0 \]
Thus,
\[ \frac{dy}{dx} = -\frac{x}{y} \]
Let this slope be \(m_1 = -\frac{x}{y}\).

For Curve 2:
\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(a^2) \]
This gives:
\[ 2x + 2y \frac{dy}{dx} = 0 \]
Thus,
\[ \frac{dy}{dx} = -\frac{x}{y} \]
Let this slope be \(m_2 = -\frac{x}{y}\).

### Step 4: Use the formula for the angle between two curves
The formula for the angle \(\theta\) between two curves with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan(\theta) = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Substituting \(m_1\) and \(m_2\):
\[ \tan(\theta) = \frac{-\frac{x}{y} - (-\frac{x}{y})}{1 + \left(-\frac{x}{y}\right)\left(-\frac{x}{y}\right)} \]
This simplifies to:
\[ \tan(\theta) = \frac{0}{1 + \frac{x^2}{y^2}} = 0 \]
Thus, \(\theta = 0\).

### Step 5: Conclusion
Since we are looking for the angle of intersection of the curves, we need to check if the angle is \(\frac{\pi}{4}\). The calculation shows that the angle is \(0\), which indicates that the curves are tangent to each other at the point of intersection, rather than intersecting at an angle of \(\frac{\pi}{4}\).