The point (-3,2) lies on a circle whose equation is (x + 3)^2 + (y + 1 )^2 =r^2. Which of the following must be the radius of the circle?

1.  3

2. A

Step-by-step explanation:

1. (-3+3)^2 + (2 + 1)^2 - r^2 = 0

0 + 9 - r^2 = 0

r^2 = 9

r = 3

2. 2x^2 + 4x + 2y^2 + 12y = - 6

Divide by 2.

(x^2 + 2x) + (y + 6y) = -3

Remember how to complete the square..

(b/2)^2

So...

(2/2) ^ 2 = 1

(6/2) ^ 2 = 9

(x^2 + 2x + 1) + (y^2 + 6y + 9) = -3 + 1 + 9 <--- You must remember to add whatever you add on the left to the right!

Now.. rewrite in standard form..  

(x+1)^2 + (y+3)^2 = 7

Careful!! the circle equation is written as

(x - h)^2 + (y - h)^2 = r^2

Meaning that the radius is square root of 7. At center (-1, -3)

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.