Area of Sector/Proof 1

Proof
Let $\mathcal C$ be a circle of radius $r$ whose center $A$ is at the origin and with radii $AB$ and $AC$ with radius $r$.

Let $B$ and $C$ be arbitrary points on the circumference of $\mathcal C$.

Let $C$ be positioned at $\tuple {0, r}$ so that $AC$ coincides with the $y$-axis.

Let $BAC$ be the sector of $\mathcal C$ whose angle between $AB$ and $AC$ is $\theta$.


 * Area-of-Sector.png

We have that $\mathcal C$ is centered at the origin.

From Equation of Circle: Corollary 2, the equation of $\mathcal C$ is given by:
 * $r^2 = x^2 + y^2$

Let $B$ have co-ordinates:
 * $\paren {x_0, y_0} = \paren {x_0, \sqrt {r^2 - {x_0}^2} }$

Triangle $\triangle ADB$ is a right triangle whose leg $DB$ is opposite $\theta$, and whose hypotenuse is $AB$.

Hence:

Consider the line $AB$, which by definition passes through $A = \paren {0, 0}$ and $B = \paren {x_0, \sqrt {r^2 - {x_0}^2} }$.

Let:
 * $\Delta y$ be the change in $y$ between $A$ and $B$
 * $\Delta x$ be the change in $x$
 * $m$ be the slope of $AB$.

Let $K$ be the region between $\mathcal C$, $AC$ and $AB$, from $0$ to $x_0$, coloured orange in the diagram above.

This is the sector of $\mathcal C$ whose area we are to find.

Let $\mathcal A$ be the area of $K$.

Then: