How do you find the equation of a circle in standard form given #x^2+y^2-10x+6y+30=0#?

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Answer 1 · 2017-01-23T09:38:58

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

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 #x^2+y^2-10x+6y+30=0# into standard form, we should group #x#-terms and #y#-terms separately as follows:

#x^2-10x+y^2+6y=-30# and now competing squares, this becomes

#(x^2-2xx5xx x+5^2)+(y^2+2xx3xxy+3^2)=-30+5^2+3^2#

or #(x-5)^2+(y+3)^2=-30+25+9#

i.e. #(x-5)^2+(y-(-3))^2=4#

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

Hence, center is #(5,-3)# and radius is #2#
graph{x^2+y^2-10x+6y+30=0 [-5, 15, -7.96, 2.04]}