Question

Let the line \( x + y = 1 \) meet the circle \( x^2 + y^2 = 4 \) at the points A and B. If the line perpendicular to AB and passing through the midpoint of the chord AB intersects the circle at C and D, then the area of the quadrilateral ABCD is equal to:

• A. \( 2\sqrt{14} \)
• B. \( 5\sqrt{7} \)
• C. \( 3\sqrt{7} \)
• D. \( \sqrt{14} \)

Hint:

To find the area of a quadrilateral formed by intersecting curves, use geometric properties and formulas for the area of triangles and rectangles.

Answer 1

The Correct Option is A

The objective is to compute the area of quadrilateral \( ACBD \), which is formed by the intersection of a circle and two lines. The process is as follows:

1. Input Data:
The circle's equation is \( x^2 + y^2 = 4 \).
The intersecting lines are \( x = y \) and \( x + y = 1 \).

2. Determination of Intersection Points:
(a) Intersection of \( x = y \) with the circle:
Substituting \( x = y \) into \( x^2 + y^2 = 4 \) yields \( x^2 + x^2 = 4 \), which simplifies to \( 2x^2 = 4 \), and further to \( x^2 = 2 \). Thus, \( x = \pm\sqrt{2} \). The intersection points are \( C (\sqrt{2}, \sqrt{2}) \) and \( D (-\sqrt{2}, -\sqrt{2}) \).

(b) Intersection of \( x + y = 1 \) with the circle:
Rearranging \( x + y = 1 \) to \( y = 1 - x \) and substituting into \( x^2 + y^2 = 4 \) gives \( x^2 + (1 - x)^2 = 4 \). Expanding this yields \( x^2 + 1 - 2x + x^2 = 4 \), which simplifies to \( 2x^2 - 2x + 1 = 4 \), and then \( 2x^2 - 2x - 3 = 0 \). Using the quadratic formula, \( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-3)}}{2(2)} = \frac{2 \pm \sqrt{4 + 24}}{4} = \frac{2 \pm \sqrt{28}}{4} = \frac{1 \pm \sqrt{7}}{2} \). The intersection points are \( A \left(\frac{1+\sqrt{7}}{2}, \frac{1-\sqrt{7}}{2}\right) \) and \( B \left(\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}\right) \).

3. Calculation of Quadrilateral \( ACBD \) Area:
Quadrilateral \( ACBD \) can be decomposed into two triangles: \( \triangle ACD \) and \( \triangle BCD \). Given that the diagonals intersect at right angles, the area of \( ACBD \) is twice the area of \( \triangle BCD \):

\( \text{Area of } ACBD = 2 \times \text{Area of } \triangle BCD \).

(a) Area of \( \triangle BCD \):
The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by \( \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \).
For vertices \( B \left(\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}\right) \), \( C (\sqrt{2}, \sqrt{2}) \), and \( D (-\sqrt{2}, -\sqrt{2}) \), the area is:

\( \text{Area of } \triangle BCD = \frac{1}{2} \left| \sqrt{2} \left(\frac{1+\sqrt{7}}{2} - (-\sqrt{2})\right) + \frac{1-\sqrt{7}}{2} \left(-\sqrt{2} - \sqrt{2}\right) + (-\sqrt{2}) \left(\sqrt{2} - \frac{1+\sqrt{7}}{2}\right) \right| \).

After simplification of the determinant expression, the area of \( \triangle BCD \) is calculated as \( \sqrt{14} \).

(b) Total Area of \( ACBD \):
Using the relationship \( \text{Area of } ACBD = 2 \times \text{Area of } \triangle BCD \), the total area is \( 2 \times \sqrt{14} = 2\sqrt{14} \).

Conclusion:
The area of the quadrilateral \( ACBD \) is \( \boxed{2\sqrt{14}} \).