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

Answer 2

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 ____.

Concept Used:

We need to find the intersection point P of the circle and parabola, determine the line L through P, find the centers of circles on y-axis that are tangent to L, and then compute the square of the area of triangle PQ₁Q₂. Key concepts include solving systems of equations, distance from a point to a line, and area of a triangle.

Step-by-Step Solution:

Step 1: Find the point of intersection P of the circle \( x^2 + y^2 = 3 \) and the parabola \( x^2 = 2y \) in the first quadrant.

Substitute \( x^2 = 2y \) from the parabola into the circle equation:

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

Since P is in the first quadrant, y > 0, so y = 1.

Then \( x^2 = 2(1) = 2 \Rightarrow x = \sqrt{2} \) (taking positive root for first quadrant).

Thus, \( P = (\sqrt{2}, 1) \).

Step 2: Find the value of α for line L passing through P.

Substitute P into the line equation \( \sqrt{2}x + y = \alpha \):

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

So the line L is \( \sqrt{2}x + y = 3 \).

Step 3: Find the centers Q₁ and Q₂ of circles on y-axis that are tangent to L.

Since centers lie on y-axis, let \( Q_1 = (0, k_1) \) and \( Q_2 = (0, k_2) \).

The circles have radius \( r = 2\sqrt{3} \) and are tangent to line L.

The distance from center (0,k) to line L equals the radius:

\[ \frac{|\sqrt{2}(0) + k - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}} = 2\sqrt{3} \] \[ \frac{|k - 3|}{\sqrt{2 + 1}} = 2\sqrt{3} \] \[ \frac{|k - 3|}{\sqrt{3}} = 2\sqrt{3} \] \[ |k - 3| = 2\sqrt{3} \cdot \sqrt{3} = 2 \cdot 3 = 6 \]

So \( k - 3 = \pm 6 \Rightarrow k = 9 \) or \( k = -3 \).

Thus, \( Q_1 = (0, 9) \) and \( Q_2 = (0, -3) \).

Step 4: Find the area of triangle PQ₁Q₂.

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

Using the formula for area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃):

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

Substitute the coordinates:

\[ \text{Area} = \frac{1}{2} |\sqrt{2}(9 - (-3)) + 0((-3) - 1) + 0(1 - 9)| \] \[ = \frac{1}{2} |\sqrt{2}(12) + 0 + 0| \] \[ = \frac{1}{2} \cdot 12\sqrt{2} = 6\sqrt{2} \]

Step 5: Find the square of the area.

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

Hence, the square of the area of triangle PQ₁Q₂ is 72.