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

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

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Answer 1 · 2016-11-18T12:20:14

The center of this circle is $(- 1, 3)$ $" " (-1,3)" "$ and radius $\frac{1}{\sqrt{5}}$ $" "1/sqrt5$

Explanation:

$" "$
The general form of the equation of a circle is :
$" "$
${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $color(blue)((x-a)^2 + (y-b)^2 = r^2)$
$" "$
where $(a, b)$ $" "(a,b)" "$ is the center of the circle and $r$ $r$ is its radius.
$" "$
$" "$
In the given exercise we are asked to transform this standard form
$" "$
into the general form of the equation of a circle.
$" "$
$" "$
First , the coefficients of $x^{2} and y^{2}$ $" "x^2 " "and" "y^2" "$ should be equal to $1$ $1$ .
$" "$
Then, complete the square.
$" "$
$" "$
$5 x^{2} + 5 y^{2} + 10 x - 30 y + 49 = 0$ $5x^2+5y^2+10x-30y+49=0$
$" "$
$\Rightarrow \frac{5 x^{2} + 5 y^{2} + 10 x - 30 y + 49}{5} = \frac{0}{5}$ $rArr(5x^2+5y^2+10x-30y+49)/5=0/5$
$" "$
$\Rightarrow x^{2} + y^{2} + 2 x - 6 y + \frac{49}{5} = 0$ $rArrx^2 + y^2 + 2x - 6y + 49/5 = 0$
$" "$
$\Rightarrow (x^{2} + 2 x + 1) + (y^{2} - 6 y + 9) + (\frac{49}{5} - 1 - 9) = 0$ $rArr(x^2 + 2x + 1) + (y^2 -6y +9) +(49/5 - 1 -9)=0$
$" "$
$\Rightarrow (x^{2} + 2 x + 1) + (y^{2} - 6 y + 9) + (\frac{49}{5} - \frac{5}{5} - \frac{45}{5}) = 0$ $rArr(x^2 +2x +1)+ (y^2-6y+9) +(49/5-5/5-45/5)=0$
$" "$
$\Rightarrow (x^{2} + 2 x + 1) + (y^{2} - 6 y + 9) + (- \frac{1}{5}) = 0$ $rArr(x^2 +2x +1)+ (y^2-6y+9) +(-1/5)=0$
$" "$
$\Rightarrow {(x + 1)}^{2} + {(y - 3)}^{2} = \frac{1}{5}$ $rArr(x +1)^2+ (y-3)^2 = 1/5$
$" "$
The equation of the circle is: ${(x + 1)}^{2} + {(y - 3)}^{2} = {(\frac{1}{\sqrt{5}})}^{2}$ $" "(x +1)^2+ (y-3)^2 = (1/sqrt5)^2$
$" "$
$" "$
Hence, The center of this circle is $(- 1, 3)$ $" " (-1,3)" "$ and radius $\frac{1}{\sqrt{5}}$ $" "1/sqrt5$

Answer 2 · 2016-11-18T12:25:06

centre $(- 1, 3)$ $(-1,3)$

radius $r = \sqrt{\frac{1}{5}} = \frac{\sqrt{5}}{5}$ $r=sqrt(1/5)=sqrt5/5$

Explanation:

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

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

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

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

Divide by $5$ $5$ first

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

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

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

$(x^{2} + 2 x + 1^{2}) + (y^{2} - 6 y + 3^{2}) - 1^{2} - 3^{2} + \frac{49}{5} = 0$ $(x^2+2xcolor(red)(+1^2))+(y^2-6ycolor(blue)(+3^2))color(red)(-1^2)color(blue)(-3^2)+49/5=0$

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

centre $(- 1, 3)$ $(-1,3)$

radius $r = \sqrt{\frac{1}{5}} = \frac{\sqrt{5}}{5}$ $r=sqrt(1/5)=sqrt5/5$

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

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Explanation:

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How do you find the center and radius for 5x^2 + 5y^2 +10x-30y+49=05x2+5y2+10x−30y+49=05x^2 + 5y^2 +10x-30y+49=0?

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Explanation:

Explanation:

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