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 :

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Aicharya · Accepted Answer

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
$\begin{aligned}
& \mathrm{A}\left(\frac{1+\sqrt{7}}{2}, \frac{1-\sqrt{7}}{2}\right) \\ & \& \mathrm{~B}\left(\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}\right)
\end{aligned}$
$\therefore$ Area of Quadrilateral ACBD
$=2 \times \text { Area of } \triangle \mathrm{BCD}$
$\begin{aligned} & =2 \times \frac{1}{2}\left|\begin{array}{ccc}\sqrt{2} & \sqrt{2} & 1 \\ \frac{1-\sqrt{7}}{2} & \frac{1+\sqrt{7}}{2} & 1 \\ -\sqrt{2} & -\sqrt{2} & 1\end{array}\right| \\ & =2 \sqrt{14}\end{aligned}$

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…

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