# 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?

• I'm not sure what you are asking Oct 12, 2018 at 14:53
• He wants to get it into a form of $y=f(x)$ Oct 12, 2018 at 14:55
• more like $x = f(y)$, but basicially the same thing. Oct 12, 2018 at 14:55
• You can't get a circle to be either $y$ as a function of $x$ or $x$ as a function of $y$, since typically for one value of the independent variable there are two values of the dependent variable. Oct 12, 2018 at 14:58
• @paw88789 but you forgot the ability to use the $\pm$ symbol. we have figured it out Oct 18, 2018 at 14:41

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: • hmmm... when i put it on desmos, only half of the circle renders. Oct 12, 2018 at 15:05
• Corrected. Thank you for pointing it out. Oct 12, 2018 at 15:08
• i hate desmos... it wont render the $\pm$ as a plus-minus, but as a variable. Oct 12, 2018 at 15:11
• wait... i can just set ± to an array with +1 and -1 as factors. then something like $d=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ becomes $d=\frac{-b+\pm\sqrt{b^2-4ac}}{2a}$ Oct 18, 2018 at 14:39