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

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 = $(6\sqrt{2})^2 = 72$