If the line `ax + y = c`, touches both the curves `x^(2) + y^(2) =1 and y^(2) = 4 sqrt2x`, then `|c|` is equal to:

Options

A. `(1)/(2)`
B. `sqrt2`
C. 2
D. `(1)/(sqrt2`

Explanation

To solve the problem, we need to find the value of \(|c|\) for the line \(ax + y = c\) that touches both the curves \(x^2 + y^2 = 1\) (a circle) and \(y^2 = 4\sqrt{2}x\) (a parabola).

### Step-by-step Solution:

1. **Identify the curves**:
- The first curve is a circle centered at the origin with radius 1: \(x^2 + y^2 = 1\).
- The second curve is a parabola given by \(y^2 = 4\sqrt{2}x\).

2. **Equation of the tangent to the parabola**:
- The standard form of the parabola is \(y^2 = 4ax\). Here, \(4a = 4\sqrt{2}\), so \(a = \sqrt{2}\).
- The equation of the tangent to the parabola in normal form is given by:
\[ y = mx + \frac{a}{m} \]
- Substituting \(a = \sqrt{2}\):
\[ y = mx + \frac{\sqrt{2}}{m} \]

3. **Distance from the center of the circle to the tangent line**:
- The distance \(d\) from the center of the circle (0,0) to the line \(y = mx + \frac{\sqrt{2}}{m}\) must equal the radius of the circle, which is 1.
- The line can be rewritten in the standard form \(mx - y + \frac{\sqrt{2}}{m} = 0\).
- The distance from the point \((0,0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|C|}{\sqrt{A^2 + B^2}} \]
- Here, \(A = m\), \(B = -1\), and \(C = \frac{\sqrt{2}}{m}\). Thus,
\[ d = \frac{\left|\frac{\sqrt{2}}{m}\right|}{\sqrt{m^2 + 1}} = 1 \]

4. **Setting up the equation**:
- From the distance formula, we have:
\[ \frac{\sqrt{2}}{|m|} = \sqrt{m^2 + 1} \]
- Squaring both sides:
\[ \frac{2}{m^2} = m^2 + 1 \]
- Multiplying through by \(m^2\):
\[ 2 = m^4 + m^2 \]
- Rearranging gives:
\[ m^4 + m^2 - 2 = 0 \]

5. **Factoring the polynomial**:
- Let \(u = m^2\). The equation becomes:
\[ u^2 + u - 2 = 0 \]
- Factoring:
\[ (u - 1)(u + 2) = 0 \]
- This gives \(u = 1\) (since \(u = -2\) is not valid for real \(m\)), so:
\[ m^2 = 1 \implies m = \pm 1 \]

6. **Finding \(c\)**:
- The tangent line equation becomes:
\[ y - mx + \frac{\sqrt{2}}{m} = 0 \]
- For \(m = 1\):
\[ y - x - \sqrt{2} = 0 \implies x - y + \sqrt{2} = 0 \implies a = -1, c = \sqrt{2} \]
- For \(m = -1\):
\[ y + x - \sqrt{2} = 0 \implies x + y - \sqrt{2} = 0 \implies a = 1, c = \sqrt{2} \]
- In both cases, we find that \(|c| = \sqrt{2}\).

### Final Answer:
\[ |c| = \sqrt{2} \]