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

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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.

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    @Jordan: If comple$t$ing squares seems to be a problem, then I suggest you look at Americo's solution.2012-01-01
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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...

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    This is honestly something I will never be able to memorize.2012-01-01
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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 $-5$).

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    @Jordan Maybe the folks here at stackexchange can help you with that too - a general plan of study or review to get comfortable with creative problem-solving ideas sounds like a great question to ask/search the archives for.2012-01-01