# 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.

• The equation for a circle with radius $r$ is going to be something like $(x-h)^2 + (y-k)^2 = r^2$. Note the $r^2$ on the right-hand side. Check that against your own work. Commented Sep 8, 2022 at 16:14

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