Step 1: Compute derivatives
The slopes of the curves at the point of intersection determine \( \tan \theta \).
1st equation: \( y^2 = x \), differentiating:
\[
2y \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{2y}.
\]
2nd equation: \( x^2 + y^2 = 2 \), differentiating:
\[
2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}.
\]
Step 2: Compute \( \tan \theta \)
\[
\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|.
\]
Substituting slopes:
\[
\tan \theta = \left| \frac{\frac{1}{2y} + \frac{x}{y}}{1 - \frac{x}{2y}} \right|.
\]
At \( x = 1, y = 1 \):
\[
\tan \theta = \left| \frac{\frac{1}{2} + 1}{1 - \frac{1}{2}} \right| = \left| \frac{\frac{3}{2}}{\frac{1}{2}} \right| = 3.
\]
Thus, the correct answer is \( \boxed{3} \).