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JEE Main 202529 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Circle

Let the line x+y=1 meet the circle x^2+y^2=4 at the points A and B...

Let the line x+y=1x+y=1 meet the circle x2+y2=4x^2+y^2=4 at the points A and B . If the line perpendicular to ABA B and passing through the mid point of the chord ABA B intersects the circle at CC and DD, then the area of the quadrilateral ADBC is equal to :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

Options

  1. A(\sqrt{14})
  2. B(3 \sqrt{7})
  3. C(2 \sqrt{14})
  4. D(5 \sqrt{7})

Correct answer

C. (2 \sqrt{14})

Step-by-step solution

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
By solving x=y\mathrm{x}=\mathrm{y} with circle We get C(2,2)D(−2,−2)\begin{aligned} & \mathrm{C}(\sqrt{2}, \sqrt{2}) \\ & \mathrm{D}(-\sqrt{2},-\sqrt{2}) \end{aligned} By solving x+y=1\mathrm{x}+\mathrm{y}=1 with circle x2+y2=4x^2+y^2=4 we set A(1+72,1−72) B(1−72,1+72)\begin{aligned} & \mathrm{A}\left(\frac{1+\sqrt{7}}{2}, \frac{1-\sqrt{7}}{2}\right) \\ & & \mathrm{~B}\left(\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}\right) \end{aligned}\thereforeAreaofQuadrilateralACBDArea of Quadrilateral ACBD=2 \times \text{ Area of } \triangle \mathrm{BCD} =2×12∣2211−721+721−2−21∣=214\begin{aligned} & =2 \times \frac{1}{2}\left|\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}\right| \\ & =2 \sqrt{14}\end{aligned}
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