How do you write the equation of the circle in standard form, identify the center and radius of #4x^2+4y^2+12x-24y+41=0#?

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Answer 1 · 2017-01-23T13:06:00

#(x-(-3/2))^2+(y-3)^2=1^2#

The standard form of equation of a circle is the center-radius form of its equation and it is in the format

#(x-h)^2+(y-k)^2=r^2#, whose center is #(h,k)# and radius is #r#.

To convert #4x^2+4y^2+12x-24y+41=0# into standard form, we should group #x#-terms and #y#-terms separately as follows:

#4x^2+12x+4y^2-24y=-41# and now competing squares, this becomes

#((2x)^2+2xx 3 xx 2x+3^2)+((2y)^2-2xx6xx2y+6^2)=-30+3^2+6^2#

or #(2x+3)^2+(2y-6)^2=-41+9+36#

i.e. #(2x+3)^2+(2y-6)^2=4# and dividing by #2^2=4#

#(x-(-3/2))^2+(y-3)^2=1^2#

Hence, center is #(-3/2,3)# and radius is #1#
graph{4x^2+4y^2+12x-24y+41=0 [-4.125, 0.875, 1.66, 4.16]}