Completing the Square to Convert a Circle Equation in General Form to Standard Form

(Video) General Form

by mathman1024

Here's a video by mathman1024 showing you how to turn an equation in general form into an equation in center-radius form so you can find the center and radius of a circle. The general form of the circle will be: $$x^2+y^2+cx+dy+e=0$$ To turn this into something we can understand we have to complete the square twice: once for the $x$ and once for the $y$.

The first problem he looks at is: $x^2+y^2+10x-14y+70=0$. The first step is to put the constant on the other side and rearrange the variable terms. $$\begin{align} &x^2+y^2+10x-14y+70=0\\ &x^2+y^2+10x-14y+70\color{#8925AE}{-70}=0\color{#8925AE}{-70}\\ &x^2+10x+y^2-14y=-70 \end{align}$$ Now we complete the square for both $x$ and $y$. Remember to do this we take the $x$ or $y$ term, half it, square it, and add the squared terms to both sides. $$\begin{align} &x^2+10x+y^2-14y=-70\\ &x^2+10x\color{#8925AE}{+\bigg(\frac{10}{2}\bigg)^2}+y^2-14y\color{#8925AE}{+\bigg(\frac{-14}{2}\bigg)^2}\\ &=-70\color{#8925AE}{+\bigg(\frac{10}{2}\bigg)^2+\bigg(\frac{-14}{2}\bigg)^2}\\ &x^2+10x+25+y^2-14y+49=4\\ &(x+5)^2+(y-7)^2=4 \end{align}$$ Now that our equation is in center-radius form we can find the center and the radius. $$\text{Center: }(-5,7)\\ r=2$$

He does one more example. Try this one on your own then see if you get the same answer.

1. $x^2+y^2-6x-4y-12=0$

Earlier, we learned about completing the square for quadratics, giving us the ability to transform equations into different forms of a quadratic. We can also use the same technique to convert a circle equation in general form to a circle equation in standard form. This technique is helpful because standard form gives us more information about the circle, including the radius and the coordinate points for the circle's center.

Recall that the general form of a circle equation is:

$$x^2+y^2+cx+dy+e=0$$

, and the standard form of a circle equation is

$$(x-h)^2 + (y-k)^2 = r^2$$

So, we are trying to go from one equation of a circle to another type:

$$x^2+y^2+cx+dy+e=0 \rightarrow (x-h)^2 + (y-k)^2 = r^2$$

Steps

Let's look at the equation we are starting with. $$x^2+y^2+cx+dy+e=0$$

Our first step is to group all like terms together. $$x^2+cx + y^2+dy+e=0$$

Next, we are going to complete the square for just the $x$ terms. We'll do this on the side, and then put the final terms back into our working equation. So, to complete the square, we have:

$(x......) - ......$

We will then look at the sign on the $x$ term, which is positive in this case. We'll add that here.

$(x +......) - ......$

Now, look at the coefficient of $x$ in the original equation. In this case, it's $c$, so we will divide $c$ by $2$. We will put this new term in the brackets.

$(x + \frac{c}{2}) - ......$

Lastly, we will square the term we just halved, and that will be the value outside of the parentheses.

$(x + \frac{c}{2}) - (\frac{c}{2})^2$

Going back, our working equation is now:

$$(x + \frac{c}{2}) - (\frac{c}{2})^2 + y^2+dy+e=0$$

We will now do the same thing for the $y$ terms. As before, we will do this work on the side, to keep all of our work straight.

$(y......) - ......$

We will then look at the sign on the $y$ term, which is positive in this case. We'll add that here.

$(y +......) - ......$

Now, look at the coefficient of $y$ in the original equation. In this case, it's $d$, so we will divide $d$ by $2$. We will put this new term in the brackets.

$(y + \frac{d}{2}) - ......$

Lastly, we will square the term we just halved, and that will be the value outside of the parentheses.

$(y + \frac{d}{2}) - (\frac{d}{2})^2$

Let's put these new terms into our working equation.

$$(x + \frac{c}{2}) - (\frac{c}{2})^2 + (y + \frac{d}{2}) - (\frac{d}{2})^2+e=0$$

Let's rearrange the equation so all of the constant terms are together.

$$(x + \frac{c}{2}) + (y + \frac{d}{2}) - (\frac{c}{2})^2 - (\frac{d}{2})^2+e=0$$

Always make sure that the right-hand side of the equation is equal to $0$. We're done! This is the basic idea of how to complete the square, transforming the equation from general form to standard form.

Completing the Square: Circles

We sometimes see a circle's equation written in general form:

$$x^2 + y^2 + Dx + Ey + F = 0$$

This form doesn't easily tell us the circle and radius - which are crucial for understanding and graphing the circle. We need to convert the equation to standard form. We can do that by completing the square.

The goal here is to group the $x$ terms and $y$ terms. Then, we can turn each group into a perfect square. For example, say you start with:

$$x^2 + 6x + y^2 - 4y = -9$$

We will group the $x$ terms and $y$ terms together.

$$(x^2 + 6x) + (y^2 - 4y) = -9$$

Now we can complete the square for each group. For the $x$ terms, take half of $6$ (which is $3$), square it (which is $9$), and add $9$. You'll do the same for the $y$ terms. You'll get $4$.

But remember - if you add $9$ and $4$ to the left side, you must do the same for the right side to keep it balanced. So, you'll have:

$$(x+3)^2 + (y -2)^2 = 4$$

The equation is now in standard form! We can easily see that the center is at $(-3,2)$ and the radius is $2$.

We can always use the center and radius to graph a circle.