Question

Let the line \( L : \sqrt{2}x + y = \alpha \) pass through the point of 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 centers \( Q_1 \) 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 1

Correct Answer: 72

Given:
\[ x^2 + y^2 = 3 \quad \text{and} \quad x^2 = 2y \]

To find the intersection point \(P\):

\[ y^2 + 2y - 3 = 0 \implies (y - 1)(y + 3) = 0 \]

Since \(y > 0\), we have:

\[ y = 1 \quad \text{and} \quad x = \sqrt{2} \implies P(\sqrt{2}, 1) \]

The line \(L : -\sqrt{2}x + y = \alpha\) passes through \(P\), so:

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

For circle \(C_1\): - Center \(Q_1\) lies on the y-axis with coordinates \((0, a)\). - Given radius \(R_1 = 2\sqrt{5}\).

Applying the condition for tangency:

\[ \left| \frac{a - 3}{\sqrt{1 + 2}} \right| = 2\sqrt{5} \]

Squaring and simplifying:

\[ |a - 3| = 6 \implies a = 9 \quad \text{or} \quad a = -3 \]

Similarly, for circle \(C_2\): - Center \(Q_2\) lies on the y-axis at \((0, -3)\).

Calculating the square of the area of triangle \(PQ_1Q_2\):

\[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} \sqrt{2} & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1 \end{vmatrix} \right| \]

\[ = \frac{1}{2} \left| \sqrt{2}(9 + 3) \right| = 6\sqrt{2} \]

Square of the area = \((6\sqrt{2})^2 = 72\)