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anonymous
 4 years ago
Find the center and the radius. x^2+y^2+4x12=0
anonymous
 4 years ago
Find the center and the radius. x^2+y^2+4x12=0

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0shoot  i just lost all my working! centre = (2,0) and radius = 4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you need the method JT?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah sorry I just dont know where to start

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok  the general equation of a circle is: (x  a)^2 + (y  b)^2 = r^2 where (a,b) is the centre and r is the radius of a circle so we need to arrange your equation to the above form x^2+y^2+4x12=0 bring terms in x ,y and numbers together x^2 + 4x + y^2  12 = 0 complete the square on x^2 + 4x: x^2 + 4x = (x + 2)^2  4  replace x^2 + 4x by (x + 2)^2  4 (x + 2)^2  4 + y^2  12 = 0 (x + 2)^2 + y^2 = 16 now put this in general form: (x + 2)^2 +( y  0)^2 = 16 compare with general form: (x  a)^2 + (y  b)^2 = r^2 x = 2, y = 0 and r^2 = 16 so r = 4 centre is (2, 0) and radius = 4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay thank you! Do you always complete the square when given an equation likte that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its one way of solving it  the other way is to use a set of formulae  i prefer the completing the square method.
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