What conic section has the equation #x^2+y^2+12x+8y=48#?

1 Answer
Sep 13, 2014

This is an equation for a circle. You begin by reorganizing the terms of the function so that #x# and #x^2# are together and #y# and #y^2# are together.

Next you will have to use the Completing the Square method.

Step 1: Reorder the terms

#x^2+12x+y^2+8y=48#

Step 2: Begin Completing the square

#x^2+12x+y^2+8y=48#

#(12/2)^2=6^2=36#, Value to be added to complete the square

#(8/2)^2=4^2=16#, Value to be added to complete the square

#x^2+12x+36+y^2+8y+16=48+36+16#

#(x^2+12x+36)+(y^2+8y+16)=100#

Factor

#(x+6)^2+(y+4)^2=100#

Solution: Standard form of a Circle.