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

Question

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 :

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

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

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

$\therefore$ 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}\end{aligned}$$\right| \
\& =2 \sqrt{14}
\]

Solutions · 1