What's the Area of the Region ?

Using the diagram above the equation of line $AC$ is $y = \dfrac{1}{2}x$ or $x = 2y$ and for the circle centered at (4,4) with have $(x - 4)^2 + (y - 4)^2 = 16$

$x = 2y \implies (2y - 4)^2 + (y - 4)^2 = 16 \implies 5y^2 - 24y + 16 = 0 \implies$

$y = \dfrac{4}{5}, y = 4$ Since we already have $(8,4) \implies P(\dfrac{8}{5},\dfrac{4}{5})$

So the area of the pink region is $A_{1} = \dfrac{1}{2}(\dfrac{4}{5})(\dfrac{8}{5}) = \boxed{\dfrac{16}{25}}$ .

$(x - 4)^2 + (y - 4)^2 = 16 \implies x = 4 - \sqrt{16 - (y - 4)^2}$ which is the portion of the circle needed.

The area of the blue region is $A_{2} = \displaystyle\int_{\dfrac{4}{5}}^{4} (4 - \sqrt{16 - (y - 4)^2}) dy$

For $I = \displaystyle\int \sqrt{16 - (y - 4)^2} dx$

Let $y - 4 = 4\sin(\theta) \implies dy = 4\cos(\theta) \implies I = 8\displaystyle\int (1 + \cos(2\theta)) d\theta$

$= 8(\theta + \sin(\theta)\cos(\theta)) = 8(\arcsin(\dfrac{y - 4}{4}) + \dfrac{(y - 4)\sqrt{16 - (y - 4)^2}}{16})$

$\implies \displaystyle\int_{\dfrac{4}{5}}^{4} (\sqrt{16 - (y - 4)^2}) dy =$ $8\arcsin(\dfrac{4}{5}) + \dfrac{96}{25}$

$\implies A_{2} = \dfrac{64}{5} - \dfrac{96}{25} - 8\arcsin(\dfrac{4}{5}) = \boxed{\dfrac{224}{25} - 8\arcsin(\dfrac{4}{5})}$

$\therefore$ The total area $A^{*} = A_{1} + A_{2} = \dfrac{48}{5} - 8\arcsin(\dfrac{4}{5}) =$ $8(6/5 - \arcsin(\dfrac{4}{5})) = 2^3(\dfrac{2 * 3}{5} - \arcsin(\dfrac{2^2}{5})) =$ $\boxed{{a^b}(\dfrac{a * b}{c} - \arcsin(\dfrac{a^{a}}{c}))}$ .

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Note: For $A_{2}$ you could have used: $$

$A_{2} = \displaystyle\int_{0}^{\dfrac{8}{5}} (4 - \sqrt{16 - (x - 4)^2} - \dfrac{4}{5}) dx =$ $\displaystyle\int_{0}^{\dfrac{8}{5}} (\dfrac{16}{5} - \sqrt{16 - (x - 4)^2}) dx$