# Finding the center and radius of a circle given a general degree 2 equation

I am trying to find the center and radius of the circle with equation $x^2 + y^2 -6x + 10y + 9 = 0$

-

Hint:
Rearrange the equation as follows: $$x^2-6x + y^2 +10y =-9$$ and complete the square in $x$ and $y$ . You should get something of the form $$(x-a)^2+(y-b)^2 = c^2$$ from which you can get the center and the radius.

All the best.

-
I forgot what that is. –  Jordan Jan 1 '12 at 21:55
I have zero idea what to do and the wikipedia link is just full of terms I don't know the meaning of, notation I don't understand and just all sorts of confusion. –  Jordan Jan 1 '12 at 22:06
@Jordan: If completing squares seems to be a problem, then I suggest you look at Americo's solution. –  jojo Jan 1 '12 at 22:34

You know the equation of a circle has the form $$r^2=(x-a)^2+(y-b)^2.$$ If you have the equation in this form, you can read off the center and radius.

But you don't have that. You need to do a bit of work to get what you have into the form you want.

The tool to use here is:

Completing the Square

To complete the square for the expression $ax^2+bx+c$ means to find constants $h$ and $k$ such that $$ax^2+bx+c=a(x-h)^2+k.$$ To do this, factor out $a$ and add and subtract $b^2\over4a^2$ in the other factor.

For your equation $$\color{maroon}{x^2} +\color{darkgreen}{ y^2} \color{maroon}{-6x }\color{darkgreen}{+ 10y }\color{maroon}{+ 9}=0$$ you want to complete the square on the $\color{maroon}x$ and on the $\color{darkgreen}y$ terms on the left hand side.

For the $y$ terms:

\eqalign{ y^2+10y &= y^2+10y +{100\over4}-{100\over4} \cr &= y^2+10y +25-25\cr &= (y+5)^2-25 }

The other part of your equation is easier to work out (note its easy when you include the 9 with it): $$x^2-6x+9=(x-3)^2$$

So

\eqalign{ & \color{maroon}{x^2} +\color{darkgreen}{ y^2} \color{maroon}{-6x }\color{darkgreen}{+ 10y }\color{maroon}{+ 9}\cr& \iff \color{maroon}{(x-3)^2}+ \color{darkgreen}{ (y+5)^2-25} =0\cr &\iff (x-3)^2+ (y+5)^2=25. }

So the center and radius are...

-
I don't understand $b^2\over4a^2$ where that comes from and why. –  Jordan Jan 1 '12 at 22:21
@Jordan $ax^2+bx =a ( x^2+{b\over a} x +D-D)$. If you want $x^2+{b\over a}x+D$ to be a perfect square, then (as it turns out) $D$ has to be $-b^2\over 4a^2$. Check by multiplying out $(x+{-b\over 2a})^2$. –  David Mitra Jan 1 '12 at 22:28
This is honestly something I will never be able to memorize. –  Jordan Jan 1 '12 at 22:29

A mechanical way is to start with the usual equation of a circle centered at the point $(a,b)$ and radius $r$ $$(x-a)^{2}+(y-b)^{2}=r^{2},\tag{1}$$ write it in the form $$(x-a)^{2}+(y-b)^{2}-r^{2}=0,$$

and expand the LHS $$\begin{eqnarray*} (x-a)^{2}+(y-b)^{2}-r^{2} &=&\left( x^{2}-2ax+a^{2}\right) +\left( y^{2}-2by+b^{2}\right) -r^{2} \\ &=&x^{2}+y^{2}-2ax-2by+a^{2}+b^{2}-r^{2}, \end{eqnarray*}$$ so that we get the equivalent equation $$x^{2}+y^{2}-2ax-2by+a^{2}+b^{2}-r^{2}=0.\tag{2}$$

Equating the coefficients of $(2)$ to the ones of the given equation $$x^{2}+y^{2}-6x+10y+9=0\tag{3}$$ results in the system of equations $$\begin{eqnarray*} -2a &=&-6 \\ -2b &=&10 \\ a^{2}+b^{2}-r^{2} &=&9, \end{eqnarray*}$$ which is equivalent to $$\begin{eqnarray*} a &=&3 \\ b &=&-5 \\ 9+25-r^{2} &=&9. \end{eqnarray*}$$ So the center is the point $(a,b)=(3,-5)$ and the radius is $r=5$, because it cannot be negative (the other solution of $9+25-r^{2} =9$ is $-5$).

-
This is a really strange solution to the problem that I doubt I would have ever seen. How did you know to do that? Did you just recognize the equation as the form of two squares involving negatives? –  Jordan Jan 1 '12 at 22:37
@Jordan I knew that the equation can be written in the form $(1)$. I just rewrote it in the form $(2)$ to compare it directely with your equation $(3)$. –  Américo Tavares Jan 1 '12 at 22:44
Is this how I should be looking at things? I never see equations like that. –  Jordan Jan 1 '12 at 22:47
@Jordan Have you never seen an equation of a circle of the form $(1)$? –  Américo Tavares Jan 1 '12 at 22:49
It has been a while but yes I have, it is just that I would not be able to translate that to (2). –  Jordan Jan 1 '12 at 22:52