How do you find the center and radius of the circle #x^2 −2x + y^2 − 6y = 26#?

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Answer 1 · 2016-05-01T06:57:54

Center is #(1,3)# and radius is #6#

General equation of a circle is of the form

#x^2+y^2+2gx+2fy+c=0#, whose center is #(-g,-f)# and radius is #sqrt(g^2+f^2-c)#

Hence in the equation of circle #x^2-2x+y^2-6y=26# is of the form

#x^2+y^2-2x-6y-26=0# and hence

#g=-1#, #f=-3# and #c=-26#

Hence, center of circle is #(-(-1),-(-3))# or #(1,3)#

and radius is #sqrt((-1)^2+(-3)^2-(-26))=sqrt(1+9+26)=sqrt36=6#

graph{x^2+y^2-2x-6y-26=0 [-10, 10, -5, 5]}