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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?

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  • $\begingroup$ I'm not sure what you are asking $\endgroup$ Oct 12, 2018 at 14:53
  • $\begingroup$ He wants to get it into a form of $y=f(x)$ $\endgroup$ Oct 12, 2018 at 14:55
  • $\begingroup$ more like $x = f(y)$, but basicially the same thing. $\endgroup$ Oct 12, 2018 at 14:55
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    $\begingroup$ 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. $\endgroup$
    – paw88789
    Oct 12, 2018 at 14:58
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    $\begingroup$ @paw88789 but you forgot the ability to use the $\pm$ symbol. we have figured it out $\endgroup$ Oct 18, 2018 at 14:41

1 Answer 1

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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:
Geogebra

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  • $\begingroup$ hmmm... when i put it on desmos, only half of the circle renders. $\endgroup$ Oct 12, 2018 at 15:05
  • $\begingroup$ Corrected. Thank you for pointing it out. $\endgroup$ Oct 12, 2018 at 15:08
  • $\begingroup$ i hate desmos... it wont render the $\pm$ as a plus-minus, but as a variable. $\endgroup$ Oct 12, 2018 at 15:11
  • $\begingroup$ 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 $d=\frac{-b+\pm\sqrt{b^2-4ac}}{2a}$ $\endgroup$ Oct 18, 2018 at 14:39

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