How do you graph #x^2+y^2+2x+2y=23#?

1 Answer
Feb 15, 2016

Answer:

Just draw a circle with center at #(-1,-1)# and radius #5#.

Explanation:

General form of quadratic equation of a conic section is #ax^2+by^2+2hxy+2gx+2fy+c=0#.

As in given the equation #x^2+y^2+2x+2y=23#, the term #xy# is not there and coefficients of #x^2# and #y^2# are equal, this is the equation of a circle.

#x^2+y^2+2x+2y=23#

#hArr# #(x^2+2x+1)+(y^2+2y+1)=23+1+1#

#hArr# #(x+1^2)+(y+1)^2=25=5^2#

As this is the equation of a circle with center at #(-1,-1)# and radius #5#.

Hence to draw the graph of #x^2+y^2+2x+2y=23#, draw a circle with center at #(-1,-1)# and radius #5#