Question

What is the center radius form of the equation of the circle

Answer 1

$x^2+y^2-2x+4y-1=0\\ x^2+y^2-2x+4y=1$

Add the square of the half of the coefficient of $x$ and $y$ to both sides.

$x^2+y^2-2x+(-1)^2+4y+(2)^2=1+(-1)^2+(2)^2\\ x^2-2x+(-1)^2+y^2+4y+(2)^2=1+1+4\\ (x-1)^2+(y+2)^2=6$

Compare this to the standard equation of a circle which is:

$$(x-a)^2+(y-b)^2=r^2$$

This implies that,

$$a=1, b=-2, r=\sqrt6$$