If common tangent to parabola `y^2=4x` and `x^2=4y` also touches the circle `x^2+y^2=c^2`, then find the value of C.

Answer

To solve the problem, we need to find the value of \( c \) such that the common tangent to the parabolas \( y^2 = 4x \) and \( x^2 = 4y \) also touches the circle \( x^2 + y^2 = c^2 \).

### Step 1: Find the equations of the common tangents to the parabolas

1. **Equation of the tangent to \( y^2 = 4x \)**:
The general equation of the tangent to the parabola \( y^2 = 4x \) is given by:
\[ y = mx + \frac{1}{m} \]
where \( m \) is the slope of the tangent.

2. **Equation of the tangent to \( x^2 = 4y \)**:
The general equation of the tangent to the parabola \( x^2 = 4y \) is given by:
\[ y = mx - \frac{m^2}{4} \]

### Step 2: Set the tangents equal to find conditions for common tangents

For the tangents to be common, their y-intercepts must be equal. Thus, we set:
\[ \frac{1}{m} = -\frac{m^2}{4} \]

### Step 3: Solve for \( m \)

Cross-multiplying gives:
\[ 4 = -m^3 \implies m^3 = -4 \implies m = -\sqrt[3]{4} \]

### Step 4: Write the equation of the common tangent

Substituting \( m = -\sqrt[3]{4} \) into the tangent equation for \( y^2 = 4x \):
\[ y = -\sqrt[3]{4}x + \frac{1}{-\sqrt[3]{4}} = -\sqrt[3]{4}x - \frac{1}{\sqrt[3]{4}} \]

### Step 5: Find the distance from the origin to the tangent line

The equation of the tangent line can be rewritten as:
\[ \sqrt[3]{4}x + y + \frac{1}{\sqrt[3]{4}} = 0 \]

To find the distance from the origin (0, 0) to this line, we use the formula for the distance from a point to a line \( Ax + By + C = 0 \):
\[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = \sqrt[3]{4} \), \( B = 1 \), and \( C = \frac{1}{\sqrt[3]{4}} \). Thus, the distance from the origin is:
\[ \text{Distance} = \frac{|\sqrt[3]{4} \cdot 0 + 1 \cdot 0 + \frac{1}{\sqrt[3]{4}}|}{\sqrt{(\sqrt[3]{4})^2 + 1^2}} = \frac{\frac{1}{\sqrt[3]{4}}}{\sqrt{(\sqrt[3]{4})^2 + 1}} \]

### Step 6: Set the distance equal to \( c \)

Since the tangent also touches the circle \( x^2 + y^2 = c^2 \), we have:
\[ c = \frac{\frac{1}{\sqrt[3]{4}}}{\sqrt{(\sqrt[3]{4})^2 + 1}} \]

### Step 7: Simplify to find \( c \)

Calculating \( c \):
1. Calculate \( (\sqrt[3]{4})^2 = \sqrt[3]{16} \).
2. Thus, \( c = \frac{1/\sqrt[3]{4}}{\sqrt{\sqrt[3]{16} + 1}} \).

After simplification, we find that:
\[ c = 1 \]

### Final Answer

The value of \( c \) is:
\[ \boxed{1} \]