Mathematics · Circles

JEE Main 2025 — 29 January, Morning Shift — Question 45

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 AB intersects is the circle at C and D , then the area of the quadrilateral ADBC is equal to

Option A:
$3 \sqrt{7}$
Option B:
$2 \sqrt{14}$
Correct
Option C:
$5 \sqrt{7}$
Option D:
$\sqrt{14}$

Answer: B

Step-by-step solution

By solving $\mathrm{x}=\mathrm{y}$ with circle We get $C(\sqrt{2}, \sqrt{2})$

$\mathrm{D}(-\sqrt{2},-\sqrt{2})$ By solving $x+y=1$ with circle $x^{2}+y^{2}=4$

we set $\mathrm{A}\left(\frac{1+\sqrt{7}}{2}, \frac{1-\sqrt{7}}{2}\right)$

$\& \mathrm{~B}\left(\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}\right)$

$\therefore$ Area of Quadrilateral ACBD $=2 \times$ Area of $\triangle \mathrm{BCD}$

$=2 \times \frac{1}{2}\left|\begin{array}{ccc}\sqrt{2} & \sqrt{2} & 1 \\ \frac{1-\sqrt{7}}{2} & \frac{1+\sqrt{7}}{2} & 1 \\ -\sqrt{2} & -\sqrt{2} & 1\end{array}\right|$ $=2 \sqrt{14}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Morning Shift
Subject: Mathematics
Chapter: Circles
Topic: Family of Circles