Considering the circumference $$x^2+y^2+2x-4y+4=0$$, find its radius and its center.
See development and solution
Development:
In this case we have $$A=-2$$, $$B=4$$ and $$C=-4$$. Therefore, the center will be $$$\Big(\dfrac{-A}{2},\dfrac{-B}{2}\Big)=\Big(\dfrac{2}{2},\dfrac{-4}{2}\Big)=(1,-2)$$$ and the radius will be $$$r=\sqrt{\Big(\dfrac{A}{2}\Big)^2+\Big(\dfrac{B}{2}\Big)^2-C}=\sqrt{\Big(\dfrac{-2}{2}\Big)^2+\Big(\dfrac{4}{2}\Big)^2+4}=$$$ $$$=\sqrt{1+4+4}=\sqrt{9}=3$$$
Solution:
Center $$(1,-2)$$ and radius $$3$$.