How would I go about graphing $2x+3 > x^2 + y^2$?

Question

How would I go about graphing this problem?

Answer 1

Hint. One may rewrite $$ x^2+y^2<2x+3 $$as $$ (x^2-2x+1)+y^2-1<3 $$or $$ (x-1)^2+y^2<4. $$

Answer 2

Hint $$4> (x^2-2x+1)+y^2 \,.$$

Answer 3

The inequality is $(x-1)^{2}+y^{2} <4$. This is simply the set of points inside the circle of radius $2$ centered at $(1,0)$.