How do you graph #x^2+y^2-4x-2y-4=0#?

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Answer 1 · 2016-02-02T09:28:15

It is an equation of a circle with center in #C=(2;1)# and radius #r=3#

From this form you can find out that the equation represents a circle.
To find its coordinates and radius you should transform it to form of:

#(x-a)^2+(y-b)^2=r^2# (1)

We start from the equation given:

#x^2+y^2-4x-2y-4=0#

Now we can group terms with the same variable:

#x^2-4x+y^2-2y-4=0#

Now we can complete the squares of variables:

#x^2-4x color(red)(+4) +y^2-2y color(red)(+1) color(red)(-5) -4=0#

I added #5# so I had to substract 5 to keep the equation balanced.

Now we can write the expressions with #x# and #y# as squares:

#(x-2)^2+(y-1)^2-9=0#

Finally I can move #9# to right side to get the form (1)

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

From this equation I can read that the equation shows a circle with center in #C=(2;1)# and radius #r=3#