How do you graph #8x ^ { 2} + 72x + 8y ^ { 2} + 16y - 342= 0#?

1 Answer
Mar 9, 2018

Answer:

It is equation of a circle with center #(-9/2,-1)# and radius #8#.

Explanation:

Observe that

  1. #8x^2+72x+8y^2+16y-342=0# is a quadratic equation,
  2. in which coefficients of #x^2# and #y^2# are equal and
  3. their is no term containing #xy#.

Hence, it represents te equation of a circle. Let us convert it to appropriate form. Diving each term by #8#, we get

#x^2+9x+y^2+2y-342/8=0# and grouping it makes it

#(x^2+9x+(9/2)^2)+(y^2+2y+1)=(9/2)^2+1+171/4#

or #(x+9/2)^2+(y+1)^2=256/4=64=8^2#

i.e. #(x+9/2)^2+(y+1)^2=8^2#

Hence, this is equation of a point whose distance from #(-9/2,-1)# is always #8# i.e.

It is equation of a circle with center #(-9/2,-1)# and radius #8#.

graph{(8x^2+72x+8y^2+16y-342)((x+9/2)^2+(y+1)^2-0.08)=0 [-20, 20, -10, 10]}