Determine the area inside the circle with centre $\left(1,-\frac{\sqrt{3}}{2}\right)$ and radius 1 that lies inside the first quadrant.
Let $C$ be the centre of the circle. Note that $C$ is in the fourth quadrant.

Let $A$ and $B$ be the points where the circle intersects the $x$-axis and let $M$ be the midpoint of $A B$.

![](https://cdn.mathpix.com/cropped/2024_04_30_dfe99ade8da3ee5d11fdg-10.jpg?height=455&width=531&top_left_y=1862&top_left_x=857)

We want to find the area of the region inside the circle and above $A B$.

We calculate this area by finding the area of sector $A C B$ and subtracting the area of $\triangle A C B$. Join $C$ to $A, B$ and $M$.

Since the radius of the circle is 1 , then $C A=C B=1$.

Since $M$ is the midpoint of $A B$ and $C$ is the centre, then $C M$ is perpendicular to $A B$, which
means that $C M$ is vertical. Since the coordinates of $C$ are $\left(1,-\frac{\sqrt{3}}{2}\right)$, then $C M=\frac{\sqrt{3}}{2}$.

Since $A C: C M=1: \frac{\sqrt{3}}{2}$, then $\triangle C A M$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, as is $\triangle B C M$.

Therefore, $\angle A C B=60^{\circ}$, and thus the area of sector $A C B$ equals $\frac{60^{\circ}}{360^{\circ}} \pi\left(1^{2}\right)=\frac{1}{6} \pi$.

Also, since $\triangle C A M$ is $30^{\circ}-60^{\circ}-90^{\circ}$, then $A M=\frac{1}{2}$. Similarly, $B M=\frac{1}{2}$.

Therefore, the area of $\triangle A C B$ is $\frac{1}{2} A B \cdot C M=\frac{1}{2}(1)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}$.

Finally, this means that the area inside the circle and inside the first quadrant is $\frac{1}{6} \pi-\frac{\sqrt{3}}{4}$.

ANSWER: $\frac{1}{6} \pi-\frac{\sqrt{3}}{4}$