Find the center and the radius. x^2+y^2+4x-12=0

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Find the center and the radius. x^2+y^2+4x-12=0

Mathematics
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shoot - i just lost all my working! centre = (-2,0) and radius = 4
do you need the method JT?
Yeah sorry I just dont know where to start

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ok - 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+4x-12=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
Okay thank you! Do you always complete the square when given an equation likte that?
its 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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