The common tangent of `x^(2)+y^(2)=4 and 2x^(2)+y^(2)=2` is

Options

A. x+y+4=0
B. x-y+7=0
C. 2x+3y+8=0
D. None of these

Answer

To find the common tangent of the given circle and ellipse, we can follow these steps:

### Step 1: Write the equations of the circle and ellipse
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This represents a circle with center at (0, 0) and radius \( r = 2 \).

The equation of the ellipse is given by:
\[ \frac{x^2}{1} + \frac{y^2}{2} = 1 \]
This can be rewritten as:
\[ x^2 + \frac{y^2}{2} = 1 \]
This represents an ellipse centered at (0, 0) with semi-major axis \( b = \sqrt{2} \) and semi-minor axis \( a = 1 \).

### Step 2: Write the general equation of the tangent lines
For the circle, the equation of the tangent line can be expressed as:
\[ y = mx \pm \sqrt{r^2(1 + m^2)} \]
Substituting \( r = 2 \):
\[ y = mx \pm 2\sqrt{1 + m^2} \]

For the ellipse, the equation of the tangent line can be expressed as:
\[ y = mx \pm \sqrt{a^2m^2 + b^2} \]
Substituting \( a = 1 \) and \( b = \sqrt{2} \):
\[ y = mx \pm \sqrt{m^2 + 2} \]

### Step 3: Set the two tangent equations equal to each other
To find the common tangents, we set the two equations equal to each other:
\[ mx + 2\sqrt{1 + m^2} = mx + \sqrt{m^2 + 2} \]
This simplifies to:
\[ 2\sqrt{1 + m^2} = \sqrt{m^2 + 2} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ 4(1 + m^2) = m^2 + 2 \]
Expanding and simplifying:
\[ 4 + 4m^2 = m^2 + 2 \]
\[ 4m^2 - m^2 = 2 - 4 \]
\[ 3m^2 = -2 \]

### Step 5: Analyze the result
Since \( m^2 \) cannot be negative, we conclude that:
\[ m^2 = -\frac{2}{3} \]
This indicates that there are no real values for \( m \).

### Conclusion
Thus, there are no common tangents to the given circle and ellipse.