Consider the parabola $C: y^{2}=4 x$ and the straight line $L: y=x+2$. Let $P$ be a variable point on $L$. Draw the two tangents from $P$ to $C$ and let $Q_{1}$ and $Q_{2}$ denote the two points of contact on $C$. Let $Q$ be the mid-point of the line segment joining $Q_{1}$ and $Q_{2}$. Find the locus of $Q$ as $P$ moves along $L$.
1. **Identify the given equations and the point \( P \):**
   - The parabola \( C \) is given by \( y^2 = 4x \).
   - The line \( L \) is given by \( y = x + 2 \).
   - Let \( P \) be a variable point on \( L \), so \( P \) has coordinates \( (\omega, \omega + 2) \).

2. **Find the tangents from \( P \) to the parabola \( C \):**
   - The general equation of the tangent to the parabola \( y^2 = 4x \) at a point \( (x_1, y_1) \) is \( yy_1 = 2(x + x_1) \).
   - Since \( P \) lies on the line \( L \), we need to find the points of tangency \( Q_1 \) and \( Q_2 \) on the parabola.

3. **Use the lemma to find the circumcenter:**
   - According to the lemma, if \( Q_1 \) and \( Q_2 \) are points of tangency, and \( X \) and \( Y \) are their projections on the directrix, then the point \( P \) is the circumcenter of \( \triangle CXY \), where \( C \) is the focus of the parabola.
   - The directrix of the parabola \( y^2 = 4x \) is \( x = -1 \), and the focus is \( (1, 0) \).

4. **Set up the equations for the tangents:**
   - The tangents from \( P \) to the parabola will intersect the parabola at points \( Q_1 \) and \( Q_2 \).
   - Let \( Q_1 = (x_1, y_1) \) and \( Q_2 = (x_2, y_2) \).

5. **Find the midpoint \( Q \) of \( Q_1Q_2 \):**
   - The coordinates of \( Q \) are \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

6. **Use the properties of the parabola and the tangents:**
   - The tangents from \( P \) to the parabola will satisfy the equation of the parabola and the tangent line equation.
   - Since \( P \) is the circumcenter, we have the relation:
     \[
     (\omega - 1)^2 + (\omega + 2)^2 = (\omega + 1)^2 + (\omega + 2 - y_2)^2 = (\omega + 1)^2 + (\omega + 2 - y_1)^2
     \]
   - Given \( Y \equiv (-1, y_1) \) and \( X \equiv (-1, y_2) \), we can use these to find the coordinates of \( Q \).

7. **Solve for the locus of \( Q \):**
   - We have \( x = \frac{x_1 + x_2}{2} \) and \( y = \frac{y_1 + y_2}{2} \).
   - Using the equation \( y - 2 = \omega \), we substitute to get:
     \[
     \omega = y - 2
     \]
   - Substitute \( \omega \) into the equation:
     \[
     \omega = -\left( x + \frac{y^2}{2} \right)
     \]
   - This simplifies to:
     \[
     y - 2 = -\left( x + \frac{y^2}{2} \right)
     \]
   - Rearrange to get the final equation:
     \[
     (y - 1)^2 = 2 \left( x - \frac{3}{2} \right)
     \]

The final answer is \( \boxed{ (y - 1)^2 = 2 \left( x - \frac{3}{2} \right) } \)