Subfield: Geometry


Question: In the diagram, the circle with centre $C(1,1)$ passes through the point $O(0,0)$, intersects the $y$-axis at $A$, and intersects the $x$-axis at $B(2,0)$. Determine, with justification, the coordinates of $A$ and the area of the part of the circle that lies in the first quadrant.

![](https://cdn.mathpix.com/cropped/2023_12_21_09a69cbcf9a187d86706g-1.jpg?height=539&width=547&top_left_y=281&top_left_x=1258)

Solution:
Since $\angle A O B=90^{\circ}, A B$ is a diameter of the circle.

Join $A B$.

![](https://cdn.mathpix.com/cropped/2023_12_21_9efb0d708494864661dbg-1.jpg?height=534&width=561&top_left_y=335&top_left_x=869)

Since $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.

Therefore, the area of the part of the circle inside the first quadrant is equal to the area of $\triangle A O B$ plus the area of the semi-circle above $A B$.

The radius of the circle is equal to the distance from $C$ to $B$, or $\sqrt{(1-2)^{2}+(1-0)^{2}}=\sqrt{2}$, so the area of the semi-circle is $\frac{1}{2} \pi(\sqrt{2})^{2}=\pi$.

The area of $\triangle A O B$ is $\frac{1}{2}(O B)(A O)=\frac{1}{2}(2)(2)=2$.

Thus, the area of the part of the circle inside the first quadrant is $\pi+2$.
Since $\angle A O B=90^{\circ}, A B$ is a diameter of the circle.

Join $A B$.

![](https://cdn.mathpix.com/cropped/2023_12_21_9efb0d708494864661dbg-1.jpg?height=540&width=551&top_left_y=1535&top_left_x=863)

Since $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.

Thus, $A O=B O$.

We "complete the square" by adding point $D(2,2)$, which is on the circle, by symmetry.



![](https://cdn.mathpix.com/cropped/2023_12_21_dded398d413ac384716fg-1.jpg?height=553&width=572&top_left_y=157&top_left_x=858)

The area of the square is 4 .

The radius of the circle is equal to the distance from $C$ to $B$, or $\sqrt{(1-2)^{2}+(1-0)^{2}}=\sqrt{2}$, so the area of the circle is $\pi(\sqrt{2})^{2}=2 \pi$.

The area of the portion of the circle outside the square is thus $2 \pi-4$. This area is divided into four equal sections (each of area $\frac{1}{4}(2 \pi-4)=\frac{1}{2} \pi-1$ ), two of which are the only portions of the circle outside the first quadrant.

Therefore, the area of the part of the circle inside the the first quadrant is $2 \pi-2\left(\frac{1}{2} \pi-1\right)=$ $\pi+2$.

Two additional ways to find the coordinates of $A$ :

$*$ The length of $O C$ is $\sqrt{1^{2}+1^{2}}=\sqrt{2}$.

Since $C$ is the centre of the circle and $O$ lies on the circle, then the circle has radius $\sqrt{2}$.

Since the circle has centre $(1,1)$ and radius $\sqrt{2}$, its equation is $(x-1)^{2}+(y-1)^{2}=2$. To find the coordinates of $A$, we substitute $x=0$ to obtain $(0-1)^{2}+(y-1)^{2}=2$ or $(y-1)^{2}=1$, and so $y=0$ or $y=2$.

Since $y=0$ gives us the point $O$, then $y=2$ gives us $A$, ie. $A$ has coordinates $(0,2)$.

* Since $O$ and $A$ are both on the circle and each has a horizontal distance of 1 from $C$, then their vertical distances from $C$ must be same, ie. must each be 1.

Thus, $A$ has coordinates $(0,2)$.

Final answer: $(0,2),\pi+2$