\( (4) \) III In the \( x y \)-plane, point \( (x, y) \) lies on the circle with equation \( \dot{x}^{2}+y^{2}=1 \) and on the line with equation \( y=2 x \). What is the value of \( x y \) ? (A) -0.45 (B) -0.40 (C) 0.40 (D) 0.45 (E) 1.00 ( \( ) \)

To find the value of \( xy \) for the point that lies on both the circle defined by the equation \( x^2 + y^2 = 1 \) and the line defined by the equation \( y = 2x \), we can substitute the equation of the line into the equation of the circle. 1. Substitute \( y = 2x \) into the circle's equation: \[ x^2 + (2x)^2 = 1 \] 2. Simplify the equation: \[ x^2 + 4x^2 = 1 \] \[ 5x^2 = 1 \] 3. Solve for \( x^2 \): \[ x^2 = \frac{1}{5} \] 4. Take the square root to find \( x \): \[ x = \pm \sqrt{\frac{1}{5}} = \pm \frac{1}{\sqrt{5}} = \pm \frac{\sqrt{5}}{5} \] 5. Now, substitute \( x \) back into the equation for \( y \): \[ y = 2x = 2\left(\pm \frac{\sqrt{5}}{5}\right) = \pm \frac{2\sqrt{5}}{5} \] 6. Now we can find \( xy \): \[ xy = x \cdot y = \left(\pm \frac{\sqrt{5}}{5}\right) \left(\pm \frac{2\sqrt{5}}{5}\right) \] 7. Calculate \( xy \): \[ xy = \frac{\sqrt{5}}{5} \cdot \frac{2\sqrt{5}}{5} = \frac{2 \cdot 5}{25} = \frac{10}{25} = \frac{2}{5} = 0.4 \] Thus, the value of \( xy \) is \( 0.4 \). The answer is \( \boxed{0.40} \).