Graph $G_1$ of a quadratic trinomial $y = px^2 + qx + r$ with real coefficients intersects the graph $G_2$ of a quadratic trinomial $y = x^2$ in points $A$, $B$. The intersection of tangents to $G_2$ in points $A$, $B$ is point $C$. If $C \in G_1$, find all possible values of $p$. 
1. **Identify the intersection points of the graphs \( G_1 \) and \( G_2 \):**
   The graphs \( G_1 \) and \( G_2 \) intersect at points \( A \) and \( B \). Therefore, the equations of the graphs are equal at these points:
   \[
   px^2 + qx + r = x^2
   \]
   Rearrange this equation to:
   \[
   (p-1)x^2 + qx + r = 0
   \]
   Let \( a \) and \( b \) be the roots of this quadratic equation. Thus, \( a \) and \( b \) are the x-coordinates of points \( A \) and \( B \), respectively.

2. **Express the roots in terms of the coefficients:**
   By Vieta's formulas, the sum and product of the roots \( a \) and \( b \) are:
   \[
   a + b = -\frac{q}{p-1}
   \]
   \[
   ab = \frac{r}{p-1}
   \]

3. **Find the equations of the tangents to \( G_2 \) at points \( A \) and \( B \):**
   The tangent to \( G_2 \) at point \( A = (a, a^2) \) is:
   \[
   y = 2ax - a^2
   \]
   Similarly, the tangent to \( G_2 \) at point \( B = (b, b^2) \) is:
   \[
   y = 2bx - b^2
   \]

4. **Determine the coordinates of point \( C \):**
   The intersection of these tangents is point \( C \). To find \( C \), solve the system of equations:
   \[
   2ax - a^2 = 2bx - b^2
   \]
   Simplify to find the x-coordinate of \( C \):
   \[
   2ax - 2bx = a^2 - b^2
   \]
   \[
   x(2a - 2b) = a^2 - b^2
   \]
   \[
   x = \frac{a^2 - b^2}{2(a - b)} = \frac{(a + b)(a - b)}{2(a - b)} = \frac{a + b}{2}
   \]
   The y-coordinate of \( C \) can be found by substituting \( x = \frac{a + b}{2} \) into either tangent equation. Using the tangent at \( A \):
   \[
   y = 2a \left( \frac{a + b}{2} \right) - a^2 = a(a + b) - a^2 = ab
   \]
   Thus, the coordinates of \( C \) are:
   \[
   C = \left( \frac{a + b}{2}, ab \right)
   \]

5. **Verify that \( C \) lies on \( G_1 \):**
   Since \( C \in G_1 \), substitute \( C \) into the equation of \( G_1 \):
   \[
   y = p \left( \frac{a + b}{2} \right)^2 + q \left( \frac{a + b}{2} \right) + r
   \]
   Given \( y = ab \), we have:
   \[
   ab = p \left( \frac{a + b}{2} \right)^2 + q \left( \frac{a + b}{2} \right) + r
   \]
   Simplify this equation:
   \[
   ab = p \left( \frac{(a + b)^2}{4} \right) + q \left( \frac{a + b}{2} \right) + r
   \]
   \[
   ab = \frac{p(a + b)^2}{4} + \frac{q(a + b)}{2} + r
   \]
   Multiply through by 4 to clear the fraction:
   \[
   4ab = p(a + b)^2 + 2q(a + b) + 4r
   \]

6. **Substitute the values of \( a + b \) and \( ab \):**
   Using \( a + b = -\frac{q}{p-1} \) and \( ab = \frac{r}{p-1} \):
   \[
   4 \left( \frac{r}{p-1} \right) = p \left( -\frac{q}{p-1} \right)^2 + 2q \left( -\frac{q}{p-1} \right) + 4r
   \]
   Simplify:
   \[
   4r = \frac{pq^2}{(p-1)^2} - \frac{2q^2}{p-1} + 4r
   \]
   Subtract \( 4r \) from both sides:
   \[
   0 = \frac{pq^2}{(p-1)^2} - \frac{2q^2}{p-1}
   \]
   Factor out \( \frac{q^2}{(p-1)^2} \):
   \[
   0 = \frac{q^2}{(p-1)^2} \left( p - 2(p-1) \right)
   \]
   Simplify inside the parentheses:
   \[
   0 = \frac{q^2}{(p-1)^2} \left( p - 2p + 2 \right)
   \]
   \[
   0 = \frac{q^2}{(p-1)^2} (2 - p)
   \]
   Since \( q^2 \neq 0 \) and \( (p-1)^2 \neq 0 \), we have:
   \[
   2 - p = 0
   \]
   Therefore:
   \[
   p = 2
   \]

The final answer is \( \boxed{ p = 2 } \)