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