Points $P$ and $Q$ are $3$ units apart. A circle centered at $P$ with a radius of $3$ units intersects a circle centered at $Q$ with radius $\sqrt{3}$ units at point $A$ and $B$. The area of the quadrilateral $APBQ$ is:
Step-by-Step Solution
Key Concept: Reflection of a point across a line and area doubling principle for symmetric quadrilaterals.
The equation of line $OB$ is $y = -\frac{3}{4}x$, which gives $4y + 3x = 0$. Point $D$ is the image of $A$ with respect to line $OB$, found using the reflection formula to get $D = \left(-\frac{24}{5}, -\frac{7}{5}\right)$. Using $\sin\theta = \frac{3}{5}$ from the altitude, we find $AD = 8$ and $OE = \frac{7}{5}$. The area of quadrilateral $APBQ$ equals $2$ times the area of $\triangle APQ$, which is calculated as $\frac{\sqrt{33}}{2} - \frac{\sqrt{99}}{2}$.
Correct Answer: 2