Solveit

yea :)

$(x-a)^2 + (y-b)^2 = r^2 \text{(equation of circle )} \\ 5x^2+3y^2-20x+24y=-53 \\ 5x^2+3y^2-20x+24y+(2)+(4)=-53+(2)+(4) \\ 5x^2-20x+2+3y^2+24y+4=-47 \\ \text{see } (a+b)^2=a^2+2ab+b^2 \text{and } (a-b)^2=a^2-2ab+b^2 \\ (\sqrt{5}*x-\sqrt{2})^2+(\sqrt{3}*x+2)^2=-53$

your equation is WRONG may be you forget smth but i did it anyway just to show you, do the same on your equation 

$(x-a)^2 + (y-b)^2 = r^2 \text{(equation of circle )}$

where your:

a=-2      (x-a)^2=(x+2)^2

b=4        (y-b)^2=(y-4)^2

r=4         r^2=16

Here s your graph:

$\pi=3.14 \\ A_{circle}= \pi r^2 \\A_{circle}=7^2 \times \pi \\ \frac{49 \pi}{6}=25.7 \text{ (aprox area of sector)}$

$\frac{11}{23}$

7000000000000000000000000

$diameter=2 \times radius$

24/ (-12)= -2