Question

For real numbers $a,b(a>b>0)$, let
Area $\{(x,y):x^2+y^2\le a^2$ and $\frac{x^2}{a^2}+\frac{y^2}{b^2}\ge 1\}=30\pi$ and
Area $\{(x,y):x^2+y^2\ge b^2$ and $\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1\}=18\pi$ Then the value of $(a-b{)}^2$ is equal to ________.

Answer 1

Answer: 12

given $\pi a^2-\pi ab=30\pi$ and $\pi ab-\pi b^2=18\pi$
on subtracting, we get $(a-b{)}^2=a^2-2ab+b^2=12$