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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

5 $\sqrt{7}$

2 $\sqrt{14}$

3 $\sqrt{7}$

$\sqrt{14}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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 $\sqrt{14}$