Given the equation #2x^2+2y^2+24x-81=0?#

Find the center and radius of a circle

1 Answer
Shwetank Mauria
Feb 24, 2017

Center is #(-6,0)# and radius is #8.746#

Explanation:

#2x^2+2y^2+24x-81=0# can be written as (dividing each term by #2#)

#x^2+y^2+12x-81/2=0#

or #x^2+2xx x xx6+6^2+y^2-6^2-81/2=0#

or #(x+6)^2+y^2=81/2+36=153/2#

or #(x-(-6))^2+(y-0)^2=(sqrt(153/2))^2#

which is locus of a point which moves so that

its distance from #(-6,0)# is always #sqrt(153/2)=8.746#

Hence, center is #(-6,0)# and radius is #8.746#
graph{2x^2+2y^2+24x-81=0 [-20, 20, -10, 10]}

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