Question

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

Answer 1

shoot - i just lost all my working!

centre = (-2,0) and radius = 4

Answer 2

do you need the method JT?

Answer 3

Yeah sorry I just dont know where to start

Answer 4

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

Answer 5

Okay thank you! Do you always complete the square when given an equation likte that?

Answer 6

its one way of solving it - the other way is to use a set of formulae - i prefer the completing the square method.