Radius & Center of Circle: Solving for x^2+4x+y^2-3y=0 | Final Review Question"

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To find the radius and center of the circle represented by the equation x^2 + 4x + y^2 - 3y = 0, one must complete the square for both the x and y terms. Completing the square for x yields (x + 2)^2 - 4, and for y yields (y - 1.5)^2 - 2.5. This reformulates the equation into the standard circle form (x - a)^2 + (y - b)^2 = r^2. The center of the circle is located at (-2, 1.5) and the radius is √2. Understanding how to manipulate the equation into this form is crucial for identifying the circle's properties.
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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 (x-a)^2+(y-b)^2=r^2, 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.
 

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