consider the general equation of a circle given by 3x^2+3y^2-12y=0

Question

A) complete the square on x and y and write this equation in the standard form

b) identify the center and radius

c) find intercepts

Answer 1

The standard form of a circle is (x-h)²+(y-k)² = r².  The center is (h,k) and the radius is r.

 

3x²+3y²-12y = 0

 

Divide by 3 to get x²+y²-4y = 0

 

                          x²+y²-4y+4 = 4

 

                          (x-0)²+(y-2)² = 2²

 

                          center:  (0,2)     radius:  2

 

x intercepts (set y = 0)   x²+4=4

 

                                       x²=0  so,x=0

 

                                       The only x-intercept is (0,0).

 

y intercepts (set x = 0)  (y-2)² = 4

 

                                   y-2 = ±2    y = 4 or y = 0

 

                                       The y-intercepts are (0,0) and (0,4).