How do you find the center and radius of the circle given #x^2+y^2+4x-8=0#?

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Answer 1 · 2016-10-22T08:02:12

we have the centre as: #(-2, 0)#

radius #sqrt(12)=2sqrt3#

The standard equation of a circle is

#(x-a)^2+(y-b)^2=r^2#

where # (a,b) # is the centre, and # r# is the radius,

we thus have to complete the square on the given equation.

#x^2+y^2+4x-8=0#

rearrange slightly

#x^2+4x+y^2-8=0#

#(x^2+4x+2^2)+y^2-8-2^2=0#

#(x+2)^2+y^2-12=0#

#(x+2)^2+y^2=12#

comparing with

#(x-a)^2+(y-b)^2=r^2#

we have the centre as: #(-2, 0)#

radius #sqrt(12)=2sqrt3#

graph{x^2+y^2+4x-8=0 [-10, 10, -5, 5]}