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 $ADBC$ is equal to
Step-by-Step Solution
Key Concept: The perpendicular bisector of chord $AB$ passes through the centre $(0,0)$ and is the line $x = y$; find all four vertices by intersection with the circle, then apply the shoelace formula on the quadrilateral $ADBC$.
The perpendicular bisector of $AB$ (along $x+y=1$, slope $-1$) through the centre $(0,0)$ is the line $x=y$. Intersecting with $x^2+y^2=4$: $C=(\sqrt{2},\sqrt{2})$, $D=(-\sqrt{2},-\sqrt{2})$. Intersecting $x+y=1$ with $x^2+y^2=4$ gives $2x^2-2x-3=0$, so $$A=\left(\tfrac{1+\sqrt{7}}{2},\tfrac{1-\sqrt{7}}{2}\right),\quad B=\left(\tfrac{1-\sqrt{7}}{2},\tfrac{1+\sqrt{7}}{2}\right).$$ Area of $ADBC$ by shoelace $= 2\times\text{Area}(\triangle BCD)$. Computing via the $3\times 3$ determinant: $$\text{Area}=2\times\tfrac{1}{2}\left|\det\begin{pmatrix}\sqrt{2}&\sqrt{2}&1\\\tfrac{1-\sqrt{7}}{2}&\tfrac{1+\sqrt{7}}{2}&1\\-\sqrt{2}&-\sqrt{2}&1\end{pmatrix}\right|=2\sqrt{14}.$$
Correct Answer: 3