Find the value of 'c' so that `2x - y + c=0` may touch the ellipse `x ^(2) + y ^(2) = 2.`

Answer

To find the value of 'c' such that the line \(2x - y + c = 0\) touches the ellipse \(x^2 + y^2 = 2\), we can follow these steps:

### Step 1: Rewrite the Ellipse in Standard Form
The given ellipse is \(x^2 + y^2 = 2\). We can rewrite this in standard form:
\[ \frac{x^2}{2} + \frac{y^2}{2} = 1 \]
From this, we can identify \(a^2 = 2\) and \(b^2 = 2\).

### Step 2: Identify the Slope of the Line
The line \(2x - y + c = 0\) can be rewritten in slope-intercept form:
\[ y = 2x + c \]
Here, the slope \(m\) of the line is \(2\).

### Step 3: Use the Tangent Condition
For a line to be tangent to an ellipse, the following condition must hold:
\[ c^2 = a^2 m^2 + b^2 \]
Substituting the values we found:
- \(a^2 = 2\)
- \(b^2 = 2\)
- \(m = 2\)

### Step 4: Substitute Values into the Tangent Condition
Now, substituting these values into the tangent condition:
\[ c^2 = 2 \cdot (2^2) + 2 \]
Calculating \(m^2\):
\[ m^2 = 2^2 = 4 \]
Now substituting:
\[ c^2 = 2 \cdot 4 + 2 \]
\[ c^2 = 8 + 2 = 10 \]

### Step 5: Solve for 'c'
Taking the square root of both sides gives:
\[ c = \pm \sqrt{10} \]

### Final Answer
Thus, the values of \(c\) for which the line \(2x - y + c = 0\) touches the ellipse \(x^2 + y^2 = 2\) are:
\[ c = \sqrt{10} \quad \text{and} \quad c = -\sqrt{10} \]

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