Question

What are the boundaries of x and y if `(x-2)^2+(y-3)^2 >=16, (x-3)^2+((y-4)^2/64)<1`?

Answer 1

The region defined by the inequations is shown in light blue.

Explanation:

`(x - 2)^2 + (y - 3)^2 ge 16` defines the exterior of a circumference centered at `{2,3}` with radius `4`

`(x - 3)^2 + (y - 4)^2/64 le 1` defines the interior of an ellipse centered at `{3,4}` having axes `1, 8`