How do you sketch the graph of #x^2+y^2+8x-6y+16=0#?

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Answer 1 · 2015-02-18T22:43:35

Hello,

Use canonical form :

#x^2+8x = (x+4)^2 - 16#.

Indeed, you have #(x+4)^4 = x^2 + 8x + 16#.

Same thing with #y# :

#y^2-6y = (y-3)^2 - 9#.

So, your equation has a good looking form :

#(x+4)^2 -16 + (y-3)^2 - 9 + 16 = 0#

or better : #(x+4)^2 + (y-3)^2 = 3^2#

You recognize the circle of center #\Omega(-4,3)#, of radius #r=3#.

graph{(x+4)^2+(y-3)^2=9 [-13.87, 6.13, -2.22, 7.78]}