Find all values of a so that the circle $x^2 - ax + y^2 + 2y = a$ has the radius 2 My goal is to find all values of "a" so that the circle $x^2 - ax + y^2 + 2y = a$ has the radius 2
The correct answer is: $a = -6$ and $a = 2$
I tried solving it by doing this:
$x^2 - ax + y^2 +2y=a$
$x^2 - ax + (y+1)^2-1=a$
$(x - \frac a2)^2 - (\frac a2)^2 + (y+1)^2-1=a$
$(x - \frac a2)^2 - {a^2\over 4} + (y+1)^2-1=a$
$(x - \frac a2)^2 + (y+1)^2=a + {a^2\over 4} + 1$
$(x - \frac a2)^2 + (y+1)^2={a^2+4a + 4\over 4}$
We want the radius to be 2 so set this ${a^2+4a + 4\over 4}$ equal to 2
${a^2+4a + 4\over 4}=2$
$a^2+4a + 4=8$
$a^2+4a -4=0$
Solve for a:
$a=-2 \pm \sqrt{4+4}$
$a=-2 \pm \sqrt{8}$
This is not correct as you can see. I don't understand what I do wrong, I'm not sure if there is one of those tiny mistakes somewhere in my solving process or if I'm completely wrong from the beginning. Thanks in advance.

$\frac{a^2+4a+4}{4}$ is not a radius.
Actually, it is the square of radius.
So, you should solve $\frac{a^2+4a+4}{4}=2^2$
And its solution is a=-6, a=2
