Let m be slope of a common tangent to the circle `x^2+y^2` = 4 and the ellipse `x^2/6+y^2/3=1`, then `m^2` equals (`sqrt2` =1.41 )

Answer

To find the value of \( m^2 \) for the slope of a common tangent to the given circle and ellipse, we can follow these steps:

### Step 1: Identify 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 a radius of 2 centered at the origin.

The equation of the ellipse is given by:
\[ \frac{x^2}{6} + \frac{y^2}{3} = 1 \]
This represents an ellipse centered at the origin with semi-major axis \( a = \sqrt{6} \) and semi-minor axis \( b = \sqrt{3} \).

### Step 2: Write the equation of the tangent to the circle
The equation of a tangent to the circle \( x^2 + y^2 = r^2 \) at a slope \( m \) is:
\[ y = mx + \sqrt{r^2(1 + m^2)} \]
For our circle, \( r = 2 \):
\[ y = mx + 2\sqrt{1 + m^2} \]

### Step 3: Write the equation of the tangent to the ellipse
The equation of a tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) at a slope \( m \) is:
\[ y = mx \pm \sqrt{a^2m^2 - b^2} \]
For our ellipse, \( a^2 = 6 \) and \( b^2 = 3 \):
\[ y = mx \pm \sqrt{6m^2 - 3} \]

### Step 4: Set the two tangent equations equal
Since both equations represent the same tangent line, we can equate them:
\[ mx + 2\sqrt{1 + m^2} = mx + \sqrt{6m^2 - 3} \]
This simplifies to:
\[ 2\sqrt{1 + m^2} = \sqrt{6m^2 - 3} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 4(1 + m^2) = 6m^2 - 3 \]
Expanding and rearranging:
\[ 4 + 4m^2 = 6m^2 - 3 \]
\[ 4 + 3 = 6m^2 - 4m^2 \]
\[ 7 = 2m^2 \]

### Step 6: Solve for \( m^2 \)
Dividing both sides by 2:
\[ m^2 = \frac{7}{2} \]

### Final Result
Thus, the value of \( m^2 \) is:
\[ \boxed{\frac{7}{2}} \]