SOLUTION: What is the center and radius of the circle with equation x^2+y^2+2x-3y-4=0

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Question 64822: What is the center and radius of the circle with equation x^2+y^2+2x-3y-4=0
Answer by funmath(2933)   (Show Source): You can put this solution on YOUR website!
What is the center and radius of the circle with equation x^2+y^2+2x-3y-4=0
Put the equation in standard form , (h,k)=center and r=radius, by completing the square.
x^2+2x+y^2-3y-4+4=0+4
x^2+2x+____+y^2-3y+_____=4+____+____
x^2+2x+(2/2)^2+y^2-3y+(-3/2)^2=4+(2/2)^2+(-3/2)^2
x^2+2x+1+y^2-3y+9/4=4+1+9/4
(x+1)^2+(y-3/2)^2=16/4+4/4+9/4
(x+1)^2+(y-3/2)^2=29/4
The center (h,k)=(-1,3/2) the radius=
Happy Calculating!!!

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