Equation of the circumference II: general equation
Find the equation of the circumference with center $(3, 4)$ and radius $2$.
$$(x-3)^2+(y-4)^2=2^2 \Rightarrow x^2+y^2-6x-8y+9+16-4=0 \Rightarrow $$ $$\Rightarrow x^2+y^2-6x-8y+21=0$$
$(x-3)^2+(y-4)^2=4$ or also $x^2+y^2-6x-8y+21=0$
Considering the circumference $x^2+y^2+2x-4y+4=0$, find its radius and its center.
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 $$r=\sqrt{\Big(\dfrac{A}{2}\Big)^2+\Big(\dfrac{B}{2}\Big)^2-C}=\sqrt{\Big(\dfrac{-2}{2}\Big)^2+\Big(\dfrac{4}{2}\Big)^2+4}=$$ $$=\sqrt{1+4+4}=\sqrt{9}=3$$
Center $(1,-2)$ and radius $3$.