How do you write the equation #-2x^2-36x-2y^2-112=0# in standard form and find the center and radius?

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Answer 1 · 2017-07-27T11:22:48

#x^2+y^2+18x+56=0#
#C=(-9;0)#
#r=5#

First, let's divide all terms of the equation by #-2# and get:

#x^2+18x+y^2+56=0#

Then let's write in standard form #x^2+y^2+ax+by+c=0#:

#x^2+y^2+18x+56=0#

where

#a=18#
#b=0#
#c=56#

Now we can get the center and radius by applying the following formulas:

#C=(-a/2;-b/2)#

#r=1/2sqrt(a^2+b^2-4c)#

Then we will get:

#C=(-cancel18^9/cancel2;0)=(-9;0)#

#r=1/2sqrt(18^2+0-4*56)=1/2sqrt100=1/2*10=5#

graph{x^2+y^2+18x+56=0 [-18, 3, -6, 6]}