Area of the circle sector

Question

Take the points $A(1, 5)$, $B(4, 4)$ and $C(-3, -3)$. All of them belong to a circumference, which center is $M$.

Consider the circle that fits perfectly inside that circumference.

Calculate the area of the circle sector restricted between the points $A$, $M$ and $B$.

Answer 1

By the Pythagoran theorem the point $(1,0)$ has distance $5$ from $A,B,C$, hence it is the circumcenter of $ABC$. Once we have $M=(1,0)$ the area of the circle sector delimited by $A,M,B$ can be computed as follows:

$$ \frac{\widehat{AMB}}{2\pi}\cdot \pi 5^2 $$ since the circumradius is $5$.
We have $\widehat{AMB}=\arctan\frac{3}{4}$ hence the wanted area is $\approx \color{red}{8.04376386}$.

Answer 2

We know that the points of a circle with a center at $(x_c, y_c)$ and radius $r$ satisfies the equation $(x-x_c)^2 + (y-y_c)^2 = r^2$. As we know three points on the circle, we can produce three simultenous equations with three unknowns. $$(1-x_c)^2+(5-y_c)^2 = r^2$$ $$(4-x_c)^2+(4-y_c)^2 = r^2$$ $$(-3-x_c)^2+(-3-y_c)^2 = r^2$$ Solving these eqautions yields $x_c=1$, $y_c=0$, $r=5$. The vector $\vec{MA}$ pointing from the center $M(1, 0)$ to $A(1, 5)$ is ${ 0 \choose 5}$, vector $\vec{MB}$ is ${3 \choose 4}$. The angle $\theta$ between these vectors can be determined using their vector product. $$ \vec{MA}\cdot\vec{MB}=0\cdot3 + 5\cdot4 = \sqrt{0^2+5^2}\cdot\sqrt{3^2+4^2}\cdot cos(\theta)$$ $$ 20 = 25\cdot cos(\theta)$$ $$\theta= arccos\left(\frac{4}{5}\right)$$ Now we can just use the formula for the area $A$ of a circular sector with radius $r$ and central angle $\theta$. $$A=\frac{r^2\theta}{2}$$ $$A=\frac{5^2\cdot\arccos\left(\frac{4}{5}\right)}{2}\approx8.04$$