Let the line x + y = 1 meet the circle x² + y² = 4 at the points A and B. If the line perpendicular to AB and passing through the mid point of the chord AB intersects is the circle at C and D, then the area of the quadrilateral ADBC is equal to

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5 $\sqrt{7}$

2 $\sqrt{14}$

3 $\sqrt{7}$

$\sqrt{14}$

answer is B.

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Detailed Solution

By solving x = y with circle
We get
$\begin{array}{l} C (\sqrt{2}, \sqrt{2}) \\ D (- \sqrt{2}, - \sqrt{2}) \end{array}$
By solving x + y = 1 with
circle x² + y² = 4
we set
$\begin{array}{l} A (\frac{1 + \sqrt{7}}{2}, \frac{1 - \sqrt{7}}{2}) \\ & B (\frac{1 - \sqrt{7}}{2}, \frac{1 + \sqrt{7}}{2}) \end{array}$
$∴$ Area of Quadrilateral ACBD
= 2 × Area of $∆$ BCD
$= 2 \times \frac{1}{2} |\begin{array}{ccc} \sqrt{2} & \sqrt{2} & 1 \\ \frac{1 - \sqrt{7}}{2} & \frac{1 + \sqrt{7}}{2} & 1 \\ - \sqrt{2} & - \sqrt{2} & 1 \end{array}|$
= 2 $\sqrt{14}$