# How do you find the center and radius for 5x^2 + 5y^2 +10x-30y+49=0?

Nov 18, 2016

The center of this circle is$\text{ " (-1,3)" }$ and radius $\text{ } \frac{1}{\sqrt{5}}$

#### Explanation:

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The general form of the equation of a circle is :
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$\textcolor{b l u e}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}}$
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where $\text{ "(a,b)" }$is the center of the circle and $r$ is its radius.
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In the given exercise we are asked to transform this standard form
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into the general form of the equation of a circle.
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First , the coefficients of $\text{ "x^2 " "and" "y^2" }$ should be equal to $1$.
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Then, complete the square.
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$5 {x}^{2} + 5 {y}^{2} + 10 x - 30 y + 49 = 0$
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$\Rightarrow \frac{5 {x}^{2} + 5 {y}^{2} + 10 x - 30 y + 49}{5} = \frac{0}{5}$
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$\Rightarrow {x}^{2} + {y}^{2} + 2 x - 6 y + \frac{49}{5} = 0$
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$\Rightarrow \left({x}^{2} + 2 x + 1\right) + \left({y}^{2} - 6 y + 9\right) + \left(\frac{49}{5} - 1 - 9\right) = 0$
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$\Rightarrow \left({x}^{2} + 2 x + 1\right) + \left({y}^{2} - 6 y + 9\right) + \left(\frac{49}{5} - \frac{5}{5} - \frac{45}{5}\right) = 0$
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$\Rightarrow \left({x}^{2} + 2 x + 1\right) + \left({y}^{2} - 6 y + 9\right) + \left(- \frac{1}{5}\right) = 0$
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$\Rightarrow {\left(x + 1\right)}^{2} + {\left(y - 3\right)}^{2} = \frac{1}{5}$
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The equation of the circle is: $\text{ } {\left(x + 1\right)}^{2} + {\left(y - 3\right)}^{2} = {\left(\frac{1}{\sqrt{5}}\right)}^{2}$
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Hence, The center of this circle is$\text{ " (-1,3)" }$ and radius $\text{ } \frac{1}{\sqrt{5}}$

Nov 18, 2016

centre $\left(- 1 , 3\right)$

radius $r = \sqrt{\frac{1}{5}} = \frac{\sqrt{5}}{5}$

#### Explanation:

To find centre and radius we need to rearrange the equation into the form:

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

where; $\left(a , b\right)$ are the co-ordinates of the centre and $r$ is the radius.

$5 {x}^{2} + 5 {y}^{2} + 10 x - 30 y + 49 = 0$

Divide by$5$ first

${x}^{2} + {y}^{2} + 2 x - 6 y + \frac{49}{5} = 0$

Now complete the square on both the $x$ and $y$ terms.

$\left({x}^{2} + 2 x\right) + \left({y}^{2} - 6 y\right) + \frac{49}{5} = 0$

$\left({x}^{2} + 2 x \textcolor{red}{+ {1}^{2}}\right) + \left({y}^{2} - 6 y \textcolor{b l u e}{+ {3}^{2}}\right) \textcolor{red}{- {1}^{2}} \textcolor{b l u e}{- {3}^{2}} + \frac{49}{5} = 0$

${\left(x + 1\right)}^{2} + {\left(y - 3\right)}^{2} = \frac{1}{5}$

centre $\left(- 1 , 3\right)$

radius $r = \sqrt{\frac{1}{5}} = \frac{\sqrt{5}}{5}$