Find $K$ such that $|(x, y)| > K$ implies $(x - 1)^2 + (y + 2)^2 > C+ 4$.

Question

For any ∈ ℝ, find such that |(, )| > ⇒ ² + ² - 2 + 4 + 1 >

i.e. ( - 1)² + ( + 2)² > + 4 whenever |(, )| >

NOTE: is a function of only, and does NOT depend on ,

What I've done: I know if |( - 1, + 2)| > $\sqrt{+4}$, then the desired inequality holds. But how do I make it so that it works with any |(x, y)| >

Answer 1

In other words, you are being asked to answer this question : given a disc, $D$, placed anywhere on the real plane, what is the radius of the largest disc centred at the origin which has a non-empty intersection with $D$. To answer this, just draw a line from the origin to the centre of $D$ and pick the farthest point(from the origin) on this line which lies in $D$.

In your question, $D$ is the disk centred at $(1, -2)$ with radius $\sqrt{C+4}$. Thus, $K = dist((0,0), (1,-2)) + \sqrt{C+4} = \sqrt{5} + \sqrt{C+4}$.