Find the Center and Radius 4x^2+48x+4y^2=0

Problem

4*x2+48*x+4*y2=0

Solution

1. Divide the entire equation by 4 to simplify the coefficients of the x2 and y2 terms.

x2+12*x+y2=0

2. Complete the square for the x terms by taking half of the coefficient of x which is 6 squaring it to get 36 and adding it to both sides of the equation.

x2+12*x+36+y2=36

3. Factor the perfect square trinomial (x2+12*x+36) into its squared binomial form.

(x+6)2+y2=36

4. Identify the center (h,k) and the radius r by comparing the equation to the standard form of a circle (x−h)2+(y−k)2=r2

h=−6

k=0

r2=36⇒r=6

Final Answer

Center: *(−6,0), Radius: *6

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