How do you graph #x^2+y^2+3x+4y+4=0#?

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How do you graph #x^2+y^2+3x+4y+4=0#?

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Answer 1 · 2016-06-04T11:18:17

The graph is the circumference

#(x-3/3)^2+(y-2)^2=(3/2)^2#

Explanation:

#x^2+y^2+3x+4y+4 = 0# can be represented in the form
#(x-x_c)^2+(y-y_c)^2-r^2=0# which is the equation of the circumference with center at #{x_c, y_c}# and radius #r#.
Equating coefficients we get

#x_c = 3/2,y_c=2,x_c^2+y_c^2-r^2=4->r=3/2#

The graph is the circumference

#(x-3/3)^2+(y-2)^2=(3/2)^2#

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How do you graph #x^2+y^2+3x+4y+4=0#?

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