How do you graph #x^2+y^2+6x-16y+48=0#?

1 Answer
Feb 1, 2016

Answer:

Rearrange and group the terms to get the expression into the standard form of one of the conics - in this case a circle. You can then identify the key points that you need to be able to graph it.

Explanation:

#x^2 + 6x +y^2 -16y +48 = 0#

#(x+3)^2 -9 + (y - 8)^2 -64 +48 = 0#

#(x+3)^2 +(y-8)^2 =73-48 = 25#

This is now recognisably a circle in the form

#(x-h)^2 +(y-k)^2 = r^2#

where #(h,k)# is the centre of the circle and #r# is the radius.

In this example #(-3,8)# is the centre and the radius is #5#