Angle of intersection of the following curves
`xy=a^2,x^2+y^2=2a^2` is 0 i.e. the touch.

Answer

To find the angle of intersection of the curves \(xy = a^2\) and \(x^2 + y^2 = 2a^2\), we will follow these steps:

### Step 1: Identify the curves
The first curve is given by:
\[ xy = a^2 \]
The second curve is given by:
\[ x^2 + y^2 = 2a^2 \]

### Step 2: Find the points of intersection
To find the points of intersection, we can solve these equations simultaneously. From the first equation, we can express \(y\) in terms of \(x\):
\[ y = \frac{a^2}{x} \]
Substituting this into the second equation:
\[ x^2 + \left(\frac{a^2}{x}\right)^2 = 2a^2 \]
This simplifies to:
\[ x^2 + \frac{a^4}{x^2} = 2a^2 \]
Multiplying through by \(x^2\) to eliminate the fraction:
\[ x^4 - 2a^2x^2 + a^4 = 0 \]
Letting \(u = x^2\), we get a quadratic equation:
\[ u^2 - 2a^2u + a^4 = 0 \]

### Step 3: Solve the quadratic equation
Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ u = \frac{2a^2 \pm \sqrt{(2a^2)^2 - 4 \cdot 1 \cdot a^4}}{2 \cdot 1} \]
This simplifies to:
\[ u = \frac{2a^2 \pm \sqrt{4a^4 - 4a^4}}{2} = \frac{2a^2 \pm 0}{2} = a^2 \]
Thus, \(u = x^2 = a^2\) implies \(x = a\) or \(x = -a\).

Substituting back to find \(y\):
- If \(x = a\), then \(y = \frac{a^2}{a} = a\).
- If \(x = -a\), then \(y = \frac{a^2}{-a} = -a\).

The points of intersection are \((a, a)\) and \((-a, -a)\).

### Step 4: Find the slopes of the tangents at the points of intersection
For the first curve \(xy = a^2\), we differentiate implicitly:
\[ y + x\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x} \]
At the point \((a, a)\):
\[ m_1 = -\frac{a}{a} = -1 \]
At the point \((-a, -a)\):
\[ m_1 = -\frac{-a}{-a} = -1 \]

For the second curve \(x^2 + y^2 = 2a^2\):
\[ 2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y} \]
At the point \((a, a)\):
\[ m_2 = -\frac{a}{a} = -1 \]
At the point \((-a, -a)\):
\[ m_2 = -\frac{-a}{-a} = -1 \]

### Step 5: Calculate the angle of intersection
Using the formula for the angle \(\theta\) between two curves:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \(m_1 = -1\) and \(m_2 = -1\):
\[ \tan \theta = \left| \frac{-1 - (-1)}{1 + (-1)(-1)} \right| = \left| \frac{0}{1 + 1} \right| = 0 \]
Thus, \(\theta = 0\).

### Conclusion
The angle of intersection of the curves \(xy = a^2\) and \(x^2 + y^2 = 2a^2\) is \(0\), indicating that the curves touch each other.

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