<s>[INST] Putting the general equation for a circle in terms of x (or y) After about a year of inactivity, I'm back!!
I've been trying to solve a series of math equations by putting them in terms of x, but I've run into a problem with the famous circle equation:

$$a = \sqrt{ ( x - b ) ^ 2 + ( y - c ) ^ 2}$$
where:
a is the radius of the circle,
b is how far to the right the circle is
c is how far upwards the circle is.

After trying for about 3 weeks, i've come up with this much so far:
$$\begin{matrix}
\text{square both sides} & a ^ 2 = ( x - b ) ^ 2 + ( y - c ) ^ 2 \\
\text{isolate term with x} & ( x - b ) ^ 2 = a ^ 2 - ( y - c ) ^ 2\\
\text{square root all terms} & | x - b | = \sqrt{ a ^ 2 - ( y - c ) ^ 2 }\\
\text{absolute value} & \sqrt{ ( x - b ) ^ 2 } = \sqrt{ a ^ 2 - ( y - c ) ^ 2 }\\
\text{dead end, trying another path} & ( x - b ) ^ 2 = ( a + ( y - c ) ) \times (a - ( y - c ) )\\
\text{}
\end{matrix}$$
After that point, it just becomes a mess of trying stuff and not getting anywhere. The reason I need it in terms of x (or y) is so I can use it for a math equation I'm doing, where I find an equation to graph every letter. Could someone help? Or at least give me a suggestive nudge?
 [/INST] Does this work:  
$a = \sqrt{ ( x - b ) ^ 2 + ( y - c ) ^ 2}$
$a^2=(x-b)^2+(y-c)^2$
$(y-c)^2=a^2-(x-b)^2$
$y-c=\pm \sqrt {a^2-(x-b)^2}$
$y=c\pm \sqrt {a^2-(x-b)^2}$
For an equation of the form $x=f(y)$
$a = \sqrt{ ( x - b ) ^ 2 + ( y - c ) ^ 2}$
$a^2=(x-b)^2+(y-c)^2$
$(x-b)^2=a^2-(y-c)^2$
$x-b=\pm\sqrt {a^2-(y-c)^2}$
$x=b\pm \sqrt {a^2-(y-c)^2}$ 
It renders on Geogebra as:
 
</s>