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Circles???????????????????????????

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Find the center and radius of x^2+y^2+8y+14=0

asked Jun 21, 2013 in PRECALCULUS by payton Apprentice

2 Answers

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The given equation is x2+y2+8y +14 = 0

x2+y2+8y+14 = 0

x2+y2+8y = - 14

Add 16  to each side

x2+ y2+ 8y +16 = -14+16

(x + 0)2+ (y+4)2= 2

Compare the above equation with given formula (x -h)2+ (y -k)2 = r2

Hence (h,k) is center and r is radius

Therefore h = 0, k = -4 and  r = √2

Therefore center (h,k) = ( 0, - 4) and radius r = √2.

answered Jun 21, 2013 by goushi Pupil
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The given circle equation is x2 + y2 + 8y  +14 = 0

Add 2 on each side

x2 +  y2 + 8y + 14 + 2 = 0 + 2

x2 + y2 + 8y + 16 = 0 + 2

(x + 0)2 + (y  -(-4))2 =  2

Compare the equation with circle formula (x - h)2 + (y - k)2 = r2

Here centre is (h , k) and radius is r

h = 0 , k  = -4 , r = √2

centre (h , k) = (0 , -4)

radius r = √2

answered Jun 21, 2013 by jouis Apprentice

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