Equations of the common tangents of the circles `x^2+ y^2 = 2a^2` and the parabola `y^2 = 8ax` are

Options

A. `y=+-(x+a)`
B. `y=+-(x+2a)`
C. `y=+-(2x+a)`
D. none

Answer

To find the equations of the common tangents of the circles \(x^2 + y^2 = 2a^2\) and the parabola \(y^2 = 8ax\), we can follow these steps:

### Step 1: Identify the equations
The equation of the circle is given by:
\[ C: x^2 + y^2 = 2a^2 \]
The equation of the parabola is given by:
\[ P: y^2 = 8ax \]

### Step 2: Find the general equation of the tangent to the parabola
The general equation of the tangent to the parabola \(y^2 = 8ax\) can be expressed as:
\[ y = mx + \frac{2a}{m} \]
where \(m\) is the slope of the tangent.

### Step 3: Rewrite the tangent equation
We can rewrite the tangent equation in a standard form:
\[ y - mx - \frac{2a}{m} = 0 \]

### Step 4: Find the distance from the center of the circle to the tangent
The center of the circle is at the origin \((0, 0)\). The distance \(d\) from the center to the tangent line is given by the formula:
\[ d = \frac{|y_0 - mx_0 - \frac{2a}{m}|}{\sqrt{1 + m^2}} \]
Substituting \(x_0 = 0\) and \(y_0 = 0\):
\[ d = \frac{|\frac{2a}{m}|}{\sqrt{1 + m^2}} \]

### Step 5: Set the distance equal to the radius of the circle
The radius of the circle is \(\sqrt{2}a\). Therefore, we set up the equation:
\[ \frac{|\frac{2a}{m}|}{\sqrt{1 + m^2}} = \sqrt{2}a \]

### Step 6: Simplify the equation
We can simplify this equation by multiplying both sides by \(\sqrt{1 + m^2}\):
\[ |\frac{2a}{m}| = \sqrt{2}a \sqrt{1 + m^2} \]
Dividing both sides by \(a\) (assuming \(a \neq 0\)):
\[ |\frac{2}{m}| = \sqrt{2} \sqrt{1 + m^2} \]

### Step 7: Square both sides
Squaring both sides gives:
\[ \frac{4}{m^2} = 2(1 + m^2) \]
Rearranging this leads to:
\[ 4 = 2m^2 + 2m^4 \]
or
\[ 2m^4 + 2m^2 - 4 = 0 \]
Dividing through by 2:
\[ m^4 + m^2 - 2 = 0 \]

### Step 8: Substitute \(u = m^2\)
Let \(u = m^2\), then we have:
\[ u^2 + u - 2 = 0 \]
Factoring gives:
\[ (u - 1)(u + 2) = 0 \]
Thus, \(u = 1\) or \(u = -2\). Since \(u = m^2\), we discard \(u = -2\) as it is not valid.

### Step 9: Find the values of \(m\)
From \(u = 1\), we have:
\[ m^2 = 1 \implies m = \pm 1 \]

### Step 10: Write the equations of the tangents
Substituting \(m = 1\) and \(m = -1\) back into the tangent equation:
1. For \(m = 1\):
\[ y = x + \frac{2a}{1} \implies y = x + 2a \]
2. For \(m = -1\):
\[ y = -x + \frac{2a}{-1} \implies y = -x + 2a \]

Thus, the equations of the common tangents are:
\[ y = x + 2a \quad \text{and} \quad y = -x + 2a \]

### Final Answer:
The equations of the common tangents are:
\[ y = x + 2a \quad \text{and} \quad y = -x + 2a \]