The general form of a circle is given as x²+y²+4x-12y+4=0 (hint *complete the square to turn it to standard form) (a) what are the coordinates of the center of the circle? (b) what is the length of the radius of the circle?

Step-by-step explanation:

Rewrite x²+y²+4x-12y+4=0  by grouping x terms first, and then y terms:

x² + 4x  + y² - 12y                                  +4=0

We have to complete the square for both x² + 4x  and  y² - 12y.  Leave some space after each:

x² + 4x                    + y² - 12y                                  +4=0

Identify the coefficient of the x term.  it is 4.  Take half of that, obtaining 2.

Square this result, obtaining 4.  Add this result (4) to x² + 4x   and then

subtract 4 immediately afterward:    

x² + 4x + 4 - 4  + y² - 12y                                  +4=0     Treat the y terms in exactly the same way:  Half of -12 is -6; the square of -6 is 36; we add 36 and then subtract 36:

x² + 4x + 4 - 4  + y² - 12y + 36  - 36                                  +4=0

Now rewrite both   x² + 4x + 4   and  y² - 12y + 36 as the squares of binomials:

                                  (x + 2)^2 - 4 + (y - 6)^2        + 4  = 0

Simplifying this, we get:

                                  (x + 2)^2 - 4 + (y - 6)^2        + 4  = 0, or

                                   (x + 2)^2  + (y - 6)^2         = 0

This indicates that the center of this circle is at (-2, 6).

But with the right side = to 0, we can only conclude that the radius of the circle is zero (0); the circle here is nothing more than a point.

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb

fdvs

step-by-step explanation: vsfdvsfvd