I am trying to find the center and radius of the circle with equation $x^2 + y^2 -6x + 10y + 9 = 0$
$\endgroup$3 Answers
$\begingroup$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.
$\endgroup$ 4 $\begingroup$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...
$\endgroup$ 3 $\begingroup$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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