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 \( \boxed{2\sqrt{14}} \).

Answer 2

Step 1: Understand the given geometry.
We are given the line equation \( x + y = 1 \) and the circle equation \( x^2 + y^2 = 4 \). The line intersects the circle at points A and B. A line perpendicular to AB passing through the midpoint of chord AB intersects the circle again at points C and D. We need to find the area of quadrilateral ABCD.

Step 2: Find the points of intersection of the line and the circle.
The equation of the line is \( x + y = 1 \), which can be rewritten as: \[ y = 1 - x. \] Substitute this into the equation of the circle \( x^2 + y^2 = 4 \): \[ x^2 + (1 - x)^2 = 4. \] Expanding and simplifying: \[ x^2 + (1 - 2x + x^2) = 4 \quad \Rightarrow \quad 2x^2 - 2x + 1 = 4 \quad \Rightarrow \quad 2x^2 - 2x - 3 = 0. \] Solving this quadratic equation using the quadratic formula: \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-3)}}{2(2)} = \frac{2 \pm \sqrt{4 + 24}}{4} = \frac{2 \pm \sqrt{28}}{4} = \frac{2 \pm 2\sqrt{7}}{4} = \frac{1 \pm \sqrt{7}}{2}. \] Thus, the x-coordinates of points A and B are \( x_1 = \frac{1 + \sqrt{7}}{2} \) and \( x_2 = \frac{1 - \sqrt{7}}{2} \), and the corresponding y-coordinates are \( y_1 = 1 - x_1 \) and \( y_2 = 1 - x_2 \). Hence, the coordinates of points A and B are: \[ A\left( \frac{1 + \sqrt{7}}{2}, \frac{1 - \sqrt{7}}{2} \right) \quad \text{and} \quad B\left( \frac{1 - \sqrt{7}}{2}, \frac{1 + \sqrt{7}}{2} \right). \] Step 3: Find the midpoint of AB and the line perpendicular to AB.
The midpoint M of chord AB has coordinates: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{1}{2}, \frac{1}{2} \right). \] The slope of line AB is given by: \[ \text{slope of AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{(1 + \sqrt{7})/2 - (1 - \sqrt{7})/2}{(1 - \sqrt{7})/2 - (1 + \sqrt{7})/2} = \frac{\sqrt{7}}{-\sqrt{7}} = -1. \] The slope of the line perpendicular to AB is the negative reciprocal, which is \( 1 \). Thus, the equation of the line passing through the midpoint M and perpendicular to AB is: \[ y - \frac{1}{2} = 1 \left( x - \frac{1}{2} \right) \quad \Rightarrow \quad y = x. \] Step 4: Find the points of intersection of this line with the circle.
Substitute \( y = x \) into the equation of the circle \( x^2 + y^2 = 4 \): \[ x^2 + x^2 = 4 \quad \Rightarrow \quad 2x^2 = 4 \quad \Rightarrow \quad x^2 = 2 \quad \Rightarrow \quad x = \pm \sqrt{2}. \] Thus, the coordinates of points C and D are \( C(\sqrt{2}, \sqrt{2}) \) and \( D(-\sqrt{2}, -\sqrt{2}) \). 
Step 5: Calculate the area of quadrilateral ABCD.
We can use the formula for the area of a quadrilateral with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4) \): \[ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right|. \] Substituting the coordinates of A, B, C, and D, we calculate the area and find that the area of quadrilateral ABCD is: \[ \boxed{2\sqrt{14}}. \]