Question

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

Answer 1

The center of this circle is`" " (-1,3)" "` and radius `" "1/sqrt5`

Explanation:

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The general form of the equation of a circle is :
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`color(blue)((x-a)^2 + (y-b)^2 = r^2)`
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where `" "(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 `" "x^2 " "and" "y^2" "` should be equal to `1`.
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Then, complete the square.
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`5x^2+5y^2+10x-30y+49=0`
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`rArr(5x^2+5y^2+10x-30y+49)/5=0/5`
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`rArrx^2 + y^2 + 2x - 6y + 49/5 = 0`
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`rArr(x^2 + 2x + 1) + (y^2 -6y +9) +(49/5 - 1 -9)=0`
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`rArr(x^2 +2x +1)+ (y^2-6y+9) +(49/5-5/5-45/5)=0`
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`rArr(x^2 +2x +1)+ (y^2-6y+9) +(-1/5)=0`
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`rArr(x +1)^2+ (y-3)^2 = 1/5`
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The equation of the circle is: `" "(x +1)^2+ (y-3)^2 = (1/sqrt5)^2`
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Hence, The center of this circle is`" " (-1,3)" "` and radius `" "1/sqrt5`

Answer 2

centre `(-1,3)`

radius `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`

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

`5x^2+5y^2+10x-30y+49=0`

Divide by`5` first

`x^2+y^2+2x-6y+49/5=0`

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

`(x^2+2x)+(y^2-6y)+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=1/5`

centre `(-1,3)`

radius `r=sqrt(1/5)=sqrt5/5`