# How do you find the center and radius of the circle x^2 + y^2 - 6x + 8y = 0?

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#### Explanation

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#### Explanation:

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8
Jul 6, 2016

Centre (3, -4)

#### Explanation:

The general equation of a circle is
${x}^{2} + {y}^{2} = {r}^{2}$

Rearranging the equation
${x}^{2} - 6 x + {y}^{2} + 8 y = 0$

Then, you would want to find the perfect squares of $x$ and $y$
remembering to balance the equation
${x}^{2} - 6 x \textcolor{red}{+ 9} + {y}^{2} + 8 y \textcolor{b l u e}{+ 16} = 0$

And because you added something that wasn't there before, you have to take it away again
${x}^{2} - 6 x \textcolor{red}{+ 9 - 9} + {y}^{2} + 8 y \textcolor{b l u e}{+ 16 - 16} = 0$

Then simplifying it to:
${\left(x - 3\right)}^{2} \textcolor{red}{- 9} + {\left(y + 4\right)}^{2} \textcolor{b l u e}{- 16} = 0$

${\left(x - 3\right)}^{2}$ comes from ${x}^{2} - 6 x + 9$
${\left(y + 4\right)}^{2}$ comes from ${y}^{2} + 8 y + 16$

If we rearrange the equation like this:
${\left(x - 3\right)}^{2} + {\left(y + 4\right)}^{2} \textcolor{red}{- 9} \textcolor{b l u e}{- 16} = 0$
And simplify it to:
${\left(x - 3\right)}^{2} + {\left(y + 4\right)}^{2} - 25 = 0$

We can determine the centre and radius by:

${\left(x - 3\right)}^{2} + {\left(y + 4\right)}^{2} - 25 = 0$
Adding 25 to both sides ${\left(x - 3\right)}^{2} + {\left(y + 4\right)}^{2} = 25$

And comparing it to
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
The radius of the equation above would be at $\left(h , k\right)$ with a radius of $r$

Therefore, your centre would have the coordinates $\left(3 , - 4\right)$ and have a radius of $\sqrt{25}$ which equals to 5

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