How do you graph #x^2+y^2+6x-2y-15=0#?

Question

5378 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-03T06:44:45

This is the equation of a circle with center #(-3,1)# and radius #5#

As in the equation #x^2+y^2+6x-2y-15=0#,

coefficients of #x^2# and #y^2# are equal and there is no term having #xy# (in other words coefficient of #xy# is #0#),

hence this is the equation of a circle.

Now #x^2+y^2+6x-2y-15=0#

#hArr(x^2+6x+9)+(y^2-2y+1)=15+9+1#

#hArr(x+3)^2+(y-1)^2=5*2#

It is therefore a circle with center #(-3,1)# and radius #5#

graph{x^2+y^2+6x-2y-15=0 [-12.42, 7.58, -3.84, 6.16]}