SOLUTION: How would I sketh the graph of this equation? x^2+y^2+4x-10y=21 I started of by doing this x^2+y^2+4x-10y=21 x^2+4x+4+y^2-10y-25=21+4-25 and got (x+2)^2+(y-5)^2=0 by this

Question 401817: How would I sketh the graph of this equation?
x^2+y^2+4x-10y=21
I started of by doing this
x^2+y^2+4x-10y=21
x^2+4x+4+y^2-10y-25=21+4-25
and got
(x+2)^2+(y-5)^2=0
by this I see that the center is (-2,5)
and the radius is 0.
How would I sketch this and is what I did right?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
How would I sketh the graph of this equation?
x^2+y^2+4x-10y=21
I started of by doing this
x^2+y^2+4x-10y=21
x^2+4x+4+y^2-10y-25=21+4-25
and got
(x+2)^2+(y-5)^2=0
by this I see that the center is (-2,5)
and the radius is 0.
How would I sketch this and is what I did right?
..
You did everything right except when you completed the square for the y-term
y^2-10y-25 this should have been y^2-10x+25. The third term is always a positive number because you are taking the square of a number. Your equation of the circle would now be (x+2)^2+(y-5)^2=50. You can now sketch the circle with center at (-2,5) and radius = sqrt(50) or 5sqrt(2)

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