JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
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 \(A B\) and passing through the mid point of the chord \(A B\) intersects the circle at \(C\) and \(D\), then the area of the quadrilateral ADBC is equal to :
- A \(\sqrt{14}\)
- B \(3 \sqrt{7}\)
- C \(2 \sqrt{14}\)
- D \(5 \sqrt{7}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{14}\)
Step-by-step Solution
Detailed explanation
By solving \(\mathrm{x}=\mathrm{y}\) with circle We get \(\begin{aligned} & \mathrm{C}(\sqrt{2}, \sqrt{2}) \\ & \mathrm{D}(-\sqrt{2},-\sqrt{2}) \end{aligned}\) By solving \(\mathrm{x}+\mathrm{y}=1\) with circle \(x^2+y^2=4\) we set…
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