How do you find the radius of the circle #x^2 + y^2 - 4x + 6y - 12 = 0#?

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Answer 1 · 2016-05-26T14:19:48

#r=5# in #(x-2)^2+(y+3)^2=5^2#

The circle equation can be arranged as

#(x-x_0)^2+(y-y_0)^2=r^2#

in which #x_0,y_0# are the center coordinates and #r# the radius.
Expanding and comparing

#x^2+y^ 2-2x_0 x-2 y_0 y +x_0^2+y_0^2-r^2 = 0#
#x^2+y^2-4x+6y-12=0#

so we have

#{(-2x_0=-4),(-2y_0=6),(x_0^2+y_0^2-r^2=-12):}#

Solving for #x_0,y_0,r# easily we obtain

#x_0=2,y_0=-3,r=5#

so the equation reads

#(x-2)^2+(y+3)^2=5^2#