JEE MAIN - Mathematics (2025 - 29th January Morning Shift - No. 17)
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 AB and passing through the mid-point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ABCD is equal to :$ \sqrt{14} $
$ 3\sqrt{7} $
$ 2\sqrt{14} $
$ 5\sqrt{7} $
Explanation
_29th_January_Morning_Shift_en_17_1.png)
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
$$\mathrm{A}\left(\frac{1+\sqrt{7}}{2}, \frac{1-\sqrt{7}}{2}\right)$$
& $\mathrm{B}\left(\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}\right)$
$\therefore$ Area of Quadrilateral ACBD
$=2 \times$ Area of $\triangle \mathrm{BCD}$
$$\begin{aligned} & =2 \times \frac{1}{2}\left|\begin{array}{ccc} \sqrt{2} & \sqrt{2} & 1 \\ \frac{1-\sqrt{7}}{2} & \frac{1+\sqrt{7}}{2} & 1 \\ -\sqrt{2} & -\sqrt{2} & 1 \end{array}\right| \\ & =2 \sqrt{14} \end{aligned}$$
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