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LaddiusMaximus
Group Title
@mertsj find the center and radius of the given circle. x^2+y^28x+12y8=0
 one year ago
 one year ago
LaddiusMaximus Group Title
@mertsj find the center and radius of the given circle. x^2+y^28x+12y8=0
 one year ago
 one year ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.0
aka complete the square (twice)
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
you can do this by converting your equation to the form (x  a)^2 + (y  b)^2 = r^2 where the center is (a,b) and radius = r you do this by a process called completing the square
 one year ago

LaddiusMaximus Group TitleBest ResponseYou've already chosen the best response.0
ok so I have (x^28x)+(y^2+12y)=8
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
ok now convert x^6  8x and y^2 + 12y to the form (x  a)^2  c
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
and (y b)^2  d
 one year ago

LaddiusMaximus Group TitleBest ResponseYou've already chosen the best response.0
(x^28x+16)+(y^2+12y+36)=8+16+36
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
x^2  8x = (x  4)^2  16
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
and y^2 + 12y = (y + 6)^2  36
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
so we get (x 4)^2 + (y + 6)^2 = 8+16+36 = 60
 one year ago

LaddiusMaximus Group TitleBest ResponseYou've already chosen the best response.0
gotcha
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
compare the above withe standard equation and you'll get what you want
 one year ago
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