Question

What is the radius for the circle given by the equation? 

Answer 1

Use:

$(a-b)^2=a^2-2ab+b^2\qquad(*)$

$x^2+y^2-6y-12=0\ \ \ \ |+12\\\\x^2+y^2-6y=12\\\\x^2+y^2-2y\cdot3=12\ \ \ \ |+3^2\\\\x^2+\underbrace{y^2-2\cdot y\cdot3+3^2}_{(*)}=12+3^2\\\\x^2+(y-3)^2=21\\\\\boxed{k=3}$