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

We are tasked with finding the area of the quadrilateral \( ACBD \) formed by the intersection of a circle and two lines. Let us proceed step by step:

1. Given Information:
The circle has the equation:

\( x^2 + y^2 = 4 \)

It is intersected by the lines:

\( x = y \) and \( x + y = 1 \).

2. Intersection Points:
(a) Solving \( x = y \) with the circle:
Substitute \( x = y \) into \( x^2 + y^2 = 4 \):

\( x^2 + x^2 = 4 \)

\( 2x^2 = 4 \)

\( x^2 = 2 \)

\( x = \pm\sqrt{2} \).

Thus, the points of intersection are:

\( C (\sqrt{2}, \sqrt{2}) \) and \( D (-\sqrt{2}, -\sqrt{2}) \).

(b) Solving \( x + y = 1 \) with the circle:
Rewrite \( x + y = 1 \) as \( y = 1 - x \), and substitute into \( x^2 + y^2 = 4 \):

\( x^2 + (1 - x)^2 = 4 \)

\( x^2 + (1 - 2x + x^2) = 4 \)

\( 2x^2 - 2x + 1 = 4 \)

\( 2x^2 - 2x - 3 = 0 \)

Solve using the quadratic formula:

\( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-3)}}{2(2)} \)

\( x = \frac{2 \pm \sqrt{4 + 24}}{4} \)

\( x = \frac{2 \pm \sqrt{28}}{4} \)

\( x = \frac{1 \pm \sqrt{7}}{2} \).

Thus, the points of intersection 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. Area of Quadrilateral \( ACBD \):
The quadrilateral \( ACBD \) can be divided into two triangles: \( \triangle ACD \) and \( \triangle BCD \). Since the diagonals of the quadrilateral 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 given vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is:

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

Substitute the coordinates of \( B \), \( C \), and \( D \):

\( B \left(\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}\right) \), \( C (\sqrt{2}, \sqrt{2}) \), \( D (-\sqrt{2}, -\sqrt{2}) \):

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

Simplify the determinant:

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

After simplifications, the area evaluates to:

\( \text{Area of } \triangle BCD = \sqrt{14} \).

(b) Total Area of \( ACBD \):
Since \( \text{Area of } ACBD = 2 \times \text{Area of } \triangle BCD \):

\( \text{Area of } ACBD = 2 \times \sqrt{14} = 2\sqrt{14} \).

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