Question

Can anyone explain how to find the equation of a circle and how to complete a square

Answer 1

@inkyvoid

Answer 2

@inkyvoyd

Answer 3

So, basically with the equation you were given earlier you need to know how to complete the square in order to write it in a form that you'll be able to find the equation of a circle. So to begin you'll want to learn how to complete the square.

Answer 4

now, generally speaking when we first learn how to complete the square it's in the context of quadratic equations and quadratic functions, which happen to be parabolas and not circles. But, for any conic section (parabola, ellipse, circle, hyperbola), you'll find completing the square generally a useful method to rearrange terms of the equation into a standard (canonical) form that makes it easy to tell certain things about the conic - say, the vertex, or foci, or directrix, radius, etc.

Answer 5

Completing the square is especially helpful for solving quadratic equations; in fact, it's more or less the first way the method is utilized. It's probably most helpful for an intuitive understanding of what's going on with any quadratic termed equation, and why it's done...

Answer 6

To try not to confuse you, I'll start with an example with only x terms. This isn't a circle (it would be a parabola), but the process for a circle is pretty straightforward as you just do it twice.

Answer 7

so, let's look at solving the equation
\(0=x^2-2x-8\)
Now, you've probably been taught to factor it, with -4 and 2, but let's look at another way to think about the problem.
Normally when we have a very basic equation, say \(x^2=4\), we can just take the square root of both sides and get \(x=\pm 2\). We have a problem when trying to take the square root of \(0=x^2-2x-8\) though. We don't really know what that looks like.

Answer 8

Here's the thing though. What if we rewrote the equation in the form \((x-a)^2=0\)? Then we could just take the square root of both sides and our equation would be solved... There is a problem with this though; I'll show you. Let's solve this equation and see what we get:
\(x-a=\pm 0\) -> \(x=a\) Well, this kind of equation only has a single solution, despite being a quadratic. You probably know a quadratic can have either 2 solutions that are real, 1 solution that is repeated (this case), or two complex solutions... In either case, our hypothetical rewriting of our equation is impossible because our original equation has two solutions and our current one only has one. So, we'll need to make things slightly more complicated. It turns out that the model \((x-a)^2=b\) works. Let's take a look why:
\((x-a)^2=b\) -> \((x-a)=\pm \sqrt{b}\) -> \(x=a \pm \sqrt{b}\). So we have two solutions; \(a+\sqrt{b}\) and \(a-\sqrt{b}\). Of course, now the problem is to find what a and b are, and roughly, that is what completing the square does.

Answer 9

the easiest way to get an idea of how to do this is to simply expand our left hand side of \((x-a)^2=b\)... we get \(x^2-2ax+a^2=b\). Now it's important to note what variables are constants and which ones are not. Keep in mind that any time we have a constant a or b, it is technically just another number. x, is also "another number", but we don't know what it is, and in particular, we're trying to solve for it. So we expect the a's and b's to become more complicated and the x term to eventually work out to just be x=blah.

Answer 10

So if we rearrange the terms it turns out that that we'll get \(x^2-2ax+a^2-b=0\). Indeed, this is a quadratic equation in the same form as our original example \(x^2-2x-8=0\).
How do we solve for a and b?
well we know that the coefficients of the terms have to match up; in other words we can set
\(-2ax=-2x\)
\(a^2-b=-8\)
The first equation tells us that a=1, and the second tells us that \(b=\pm 3\).

Answer 11

correction: b=9; apologies

Answer 12

It turns out that because I worked backwards from solution to problem, we already have the solution. It was stated earlier as \(a+\sqrt{b},a-\sqrt{b}\) so we have 1+3 and 1-3 so the solution to this quadratic is 4 and -2.

Answer 13

So where did I complete the square? Well, I didn't... I more or less went backwards in this situation, because it makes more sense. To complete the square, you have to go in the REVERSE order.

Answer 14

you see, I started out with a "square" (x-a)^2 and then uncompleted it after showing what the solution was. To complete the square, you have to start out with a general quadratic \(ax^2+bx+c\) and then turn it INTO the form \((x-\alpha)^2+\beta \) where \(\alpha\) and \(\beta\) are constants.

Answer 15

I'll do it for the most general quadratic, then on a more specific example.
The general quadratic equation is
\(ax^2+bx+c=0\)
The leading a coefficient is a pain so we'll divide both sides by a before starting.
\(x^2+\frac{b}{a}x+\frac{c}{a}=0\)
I want something in the form \((x+\alpha)^2+\beta=0\)...
Let's expand that for an idea...
\(x^2+2\alpha x+\alpha^2+\beta=0\)
so \(2\alpha x=\frac{b}{a}x\)
and \(\alpha^2+\beta=\frac{c}{a}\)
alpha is pretty easy to solve for; it'll just be \(\alpha=\frac{b}{2a}\)
Let's nto worry about beta just yet and instead just plug in what we know for alpha:
\((x+\frac{b}{2a})^2+\beta=0\)
and let's expand and match with our original to solve for beta:
\(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}+\beta=0\)
so \(\frac{b^2}{4a^2}+\beta=\frac{c}{a}\)
well then beta must simply be \(\beta=-\frac{b^2}{4a^2}+\frac{c}{a}\)
Let's combine the terms under a single fraction; this will make things more elegant later.
\(\beta=-\frac{b^2+4ac}{4a^2}\)
So then,
\((x-\frac{b}{a})^2-\frac{b^2+4ac}{4a^2}=0\)
Yeah, the constants are very messy, but we're trying to come up with a formula here so we expect it.

Answer 16

Correction: I should've gotten \((x+\frac{b}{2a})^2-\frac{b^2+4ac}{4a^2}=0\)
So indeed, we have completed the square. Of course, stopping here is a bit silly since we're so close to getting the actual quadratic formula, so I'll go ahead and finish it:
Let's move the constant over and take the square root:
\((x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\)
\(x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\)
\(x=-\frac{b}{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}}\)
and simplifying:
\(x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\)
or
\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
which you can cross check is indeed the quadratic formula.

Answer 17

let's do a quick example.
\(x^2+4x-12=0\)
\((x+?)^2+??=0\)
we can guess that ? will be 2 because \((x+a)^2=x^2+2ax+a^2\) which gives us 2a or 2*2=4 which is the correct term. In general, when you complete the square it'll always be half of the first degree (the term with just x) term.
so...
\(((x+2)^2-4)-12=0\)
Here I did a few steps in one; this is generally the case with completing the square so take your time and see what was done.
\((x+2)^2=16\)
So here I combined the -12 and -4 and moved them over to the right hand side.
\(x+2=\pm 4\)
\(x=2\pm 4\)
or
\(x=6, x=-2\)
Well, we now have enough knowledge for one of your circle problems.

Answer 18

Feel free to post a question regarding circles since we complete the square twice, and the process and way we want to rewrite the equation is slightly different.

Answer 19

lol it's all good im just taking in the info because it's ALOT btw thank soo much for this

Answer 20

Another method for completing the square

\(ax^2+bx+c=a(x+\frac{b}{2a})^2+c-\dfrac{b^2}{4a}\)