Let the line `L : sqrt(2)x + y =alpha` pass through the point of the intersection P (in the first quadrant) of the circle `x^2 + y^2 = 3` and the parabola `x^2 = 2y`. Let the line L touch two circles `C_1` and `C_2` of equal radius `2 sqrt 3` . If the centres `Q_1` and `Q_2` of the circles `C_1` and `C_2` lie on the y-axis, then the square of the area of the triangle `PQ_1Q_2` is equal to ___________.

Answer

To solve the problem step by step, we will follow the outlined process to find the square of the area of triangle \( PQ_1Q_2 \).

### Step 1: Find the intersection point \( P \) of the circle and the parabola.

We have the equations:
1. Circle: \( x^2 + y^2 = 3 \)
2. Parabola: \( x^2 = 2y \)

Substituting \( y = \frac{x^2}{2} \) from the parabola into the circle's equation:

\[ x^2 + \left(\frac{x^2}{2}\right)^2 = 3 \]

This simplifies to:

\[ x^2 + \frac{x^4}{4} = 3 \]

Multiplying through by 4 to eliminate the fraction:

\[ 4x^2 + x^4 = 12 \]

Rearranging gives:

\[ x^4 + 4x^2 - 12 = 0 \]

Let \( z = x^2 \), then we have a quadratic equation:

\[ z^2 + 4z - 12 = 0 \]

Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ z = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm 8}{2} \]

Calculating the roots:

\[ z = 2 \quad (\text{since } z = -6 \text{ is not valid}) \]

Thus, \( x^2 = 2 \) gives \( x = \sqrt{2} \) (in the first quadrant). Now substituting back to find \( y \):

\[ y = \frac{x^2}{2} = \frac{2}{2} = 1 \]

So, the point \( P \) is:

\[ P = (\sqrt{2}, 1) \]

### Step 2: Find \( \alpha \) for line \( L \).

The line \( L \) is given by:

\[ \sqrt{2}x + y = \alpha \]

Substituting the coordinates of point \( P \):

\[ \sqrt{2}(\sqrt{2}) + 1 = \alpha \implies 2 + 1 = \alpha \implies \alpha = 3 \]

### Step 3: Determine the centers \( Q_1 \) and \( Q_2 \) of the circles.

The circles \( C_1 \) and \( C_2 \) have equal radius \( 2\sqrt{3} \) and their centers lie on the y-axis, so we can denote their centers as \( Q_1(0, \beta_1) \) and \( Q_2(0, \beta_2) \).

### Step 4: Use the distance from the center to the line.

The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For our line \( L: \sqrt{2}x + y - 3 = 0 \) (where \( A = \sqrt{2}, B = 1, C = -3 \)) and the centers \( Q_1(0, \beta_1) \) and \( Q_2(0, \beta_2) \):

\[ d = \frac{|\sqrt{2}(0) + 1(\beta) - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}} = \frac{|\beta - 3|}{\sqrt{3}} \]

Setting this equal to the radius \( 2\sqrt{3} \):

\[ \frac{|\beta - 3|}{\sqrt{3}} = 2\sqrt{3} \]

Multiplying both sides by \( \sqrt{3} \):

\[ |\beta - 3| = 6 \]

This gives two cases:

1. \( \beta - 3 = 6 \) \( \Rightarrow \beta = 9 \)
2. \( \beta - 3 = -6 \) \( \Rightarrow \beta = -3 \)

Thus, the centers are:

\[ Q_1(0, -3) \quad \text{and} \quad Q_2(0, 9) \]

### Step 5: Calculate the area of triangle \( PQ_1Q_2 \).

Using the formula for the area of a triangle given vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting \( P(\sqrt{2}, 1), Q_1(0, -3), Q_2(0, 9) \):

\[ \text{Area} = \frac{1}{2} \left| \sqrt{2}(-3 - 9) + 0(9 - 1) + 0(1 + 3) \right| \]

This simplifies to:

\[ = \frac{1}{2} \left| \sqrt{2}(-12) \right| = \frac{1}{2} \cdot 12\sqrt{2} = 6\sqrt{2} \]

### Step 6: Find the square of the area.

The square of the area is:

\[ \text{Area}^2 = (6\sqrt{2})^2 = 36 \cdot 2 = 72 \]

Thus, the final answer is:

\[ \boxed{72} \]