The slope of one of the common tangent to circle `x^(2)+y^(2)=1` and ellipse `x^(2)/4+2y^(2)=1` is `sqrt(a/b)` where `gcd(a, b)=1` then `(a+b)/2` is equal to

Answer

To find the slope of one of the common tangents to the circle \(x^2 + y^2 = 1\) and the ellipse \(\frac{x^2}{4} + 2y^2 = 1\), we can follow these steps:

### Step 1: Identify the equations of the circle and the ellipse.
The equation of the circle is:
\[ x^2 + y^2 = 1 \]
The equation of the ellipse can be rewritten as:
\[ \frac{x^2}{4} + \frac{y^2}{\frac{1}{2}} = 1 \]
This shows that \(a^2 = 4\) and \(b^2 = \frac{1}{2}\).

### Step 2: Write the equation of the tangent line.
The equation of the tangent line can be expressed in the slope-intercept form as:
\[ y = mx + c \]
where \(m\) is the slope and \(c\) is the y-intercept.

### Step 3: Use the condition for tangents to the circle.
For the circle, the condition for the line \(y = mx + c\) to be a tangent is:
\[ c^2 = r^2(1 + m^2) \]
where \(r\) is the radius of the circle. Here, \(r = 1\), thus:
\[ c^2 = 1(1 + m^2) = 1 + m^2 \]

### Step 4: Use the condition for tangents to the ellipse.
For the ellipse, the condition for the line \(y = mx + c\) to be a tangent is:
\[ \frac{c^2}{b^2} + \frac{c^2}{a^2} = 1 \]
Substituting \(a^2 = 4\) and \(b^2 = \frac{1}{2}\):
\[ \frac{c^2}{\frac{1}{2}} + \frac{c^2}{4} = 1 \]
This simplifies to:
\[ 2c^2 + \frac{c^2}{4} = 1 \]
Multiplying through by 4 to eliminate the fraction:
\[ 8c^2 + c^2 = 4 \implies 9c^2 = 4 \implies c^2 = \frac{4}{9} \]

### Step 5: Equate the expressions for \(c^2\).
From the circle:
\[ c^2 = 1 + m^2 \]
From the ellipse:
\[ c^2 = \frac{4}{9} \]
Setting these equal gives:
\[ 1 + m^2 = \frac{4}{9} \]
Rearranging gives:
\[ m^2 = \frac{4}{9} - 1 = \frac{4}{9} - \frac{9}{9} = -\frac{5}{9} \]
This is incorrect, so we must have made an algebraic error. Let's go back to the ellipse condition.

### Step 6: Correctly set up the ellipse condition.
The correct condition for the ellipse is:
\[ \frac{c^2}{\frac{1}{2}} + \frac{c^2}{4} = 1 \]
This simplifies to:
\[ 2c^2 + \frac{c^2}{4} = 1 \implies 8c^2 + c^2 = 4 \implies 9c^2 = 4 \implies c^2 = \frac{4}{9} \]

### Step 7: Substitute \(c^2\) back into the circle condition.
Now substituting \(c^2 = \frac{4}{9}\) into the circle condition:
\[ \frac{4}{9} = 1 + m^2 \implies m^2 = \frac{4}{9} - 1 = \frac{4}{9} - \frac{9}{9} = -\frac{5}{9} \]
This is still incorrect. Let's go back to the tangent equations.

### Step 8: Solve for \(m\).
We need to equate the two conditions correctly. From the ellipse:
\[ \frac{c^2}{\frac{1}{2}} + \frac{c^2}{4} = 1 \implies 2c^2 + \frac{c^2}{4} = 1 \]
This leads to:
\[ \frac{9c^2}{4} = 1 \implies c^2 = \frac{4}{9} \]
Now substituting back into the circle:
\[ \frac{4}{9} = 1 + m^2 \implies m^2 = \frac{4}{9} - 1 = \frac{4}{9} - \frac{9}{9} = -\frac{5}{9} \]
This is incorrect.

### Step 9: Find the slope \(m\).
Going back to the ellipse:
\[ c^2 = 4m^2 + \frac{1}{2} \]
Equating gives:
\[ 1 + m^2 = 4m^2 + \frac{1}{2} \]
This simplifies to:
\[ 1 - \frac{1}{2} = 4m^2 - m^2 \implies \frac{1}{2} = 3m^2 \implies m^2 = \frac{1}{6} \]

### Step 10: Find \(a\) and \(b\).
Here, \(m = \pm \sqrt{\frac{1}{6}}\). Thus, we have \(a = 1\) and \(b = 6\). The GCD of \(1\) and \(6\) is \(1\).

### Step 11: Calculate \((a + b)/2\).
Now, we calculate:
\[ \frac{a + b}{2} = \frac{1 + 6}{2} = \frac{7}{2} \]

### Final Answer:
Thus, the final answer is:
\[ \frac{7}{2} \]