Practise Calc Question

Question

How is the radius $10$ in the circle equation: $x^2+y^2+6x-4y+3=0$? My work: Standard Form for circle: $(x-a)^2+(y-b)^2=r^2$

$X:$
$X^2+6x+[?]=-3+[?]$
$(x+3)^2= X^2+6x+9=-3+9 X^2+6x+9=6$

$Y:$
$Y^2-4y+[?]=-3+[?] $
$(y-2)^2= y^2-4y+4=-3+4 y^2-4y+4=-1$
The standard form of the Circle is: $(x+3)^2+(y-2)^2=r^2$ My question is: How do you find the radius of the circle?

Answer 1

We start with\begin{align}x^2+y^2+6x-4y+3&=0 \\ x^2+6x+y^2-4y&=-3 \\x^2+6x+9+y^2-4y+4&=-3+9+4\\(x+3)^2+(y-2)^2&=10 \end{align} which is of the standard form $(x-h)^2+(y-k)^2=r^2$, where $r$ is the radius and $(h,k)$ is the center of the circle.

Notice that $r^2=10$. Taking square root on both sides, we get $r = \pm \sqrt{10}$. But the radius is an absolute value, so we can omit the negative solution. So we get $\boxed{r=\sqrt{10}}$.

Answer 2

You have the good idea : $$x^2+6x=(x+3)^2-3^2\\ y^2 -4y =(y-2)^2-2^2 $$ so $$x^2+y^2+6x-4y+3=(x+3)^2-3^2+(y-2)^2-2^2 +3$$ From that, I let you find $r$.

Answer 3

You're given $$x^2+6x + y^2 - 4y = -3$$ Complete the square in $x$ to get $$(x^2+6x+9) + y^2-4y = -3 + 9$$ Now complete the square in $y$ to get $$(x^2 + 6x + 9) + (y^2 - 4y + 4) = -3 + 9 + 4$$ And with a little arithmetic, we conclude: $$(x+3)^2 + (y-2)^2 = 10 = r^2$$ and this immediately gives you the radius.

One way to do this is to get all the $x^2$, $x$, $y^2$, and $y$ terms on one side and then to get the number on the other side. You then complete the square exactly as you did (at each step making sure that whatever you add to one side, you add the same thing to the other side in order to preserve equality).