Finding radius and center of x^2+4x+y^2-3y=0

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imdapolak
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1. What are the radius and center of this circle? x^2+4x+y^2-3y=0



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3. Another question on my final review I don't recall exactly how to set up or what steps I need to do in order to solve this problem. Any help is appreciated
 
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If I gave you the equation in the form [itex](x-a)^2+(y-b)^2=r^2[/itex], could you tell me the location of the center of the circle and its radius?

If so, just complete the square for both x and y in your above equation in order to get it into this form.
 
The equation of a circle is of the form (x-a)^2+(y-b)^2=r^2. So you want to write x^2+4x as (x-a)^2+constant. In other words complete the square.