Question

circle a (shown) is defined by the equation $(x + 2)^2 + y^2 = 9$. circle b (not shown) is the result of shifting circle a down 6 units and increasing the radius so that the radius of circle b is 2 times the radius of circle a. which equation defines circle b?
a $(x + 2)^2+(y + 6)^2=(4)(9)$
b $2(x + 2)^2+2(y + 6)^2 = 9$
c $(x + 2)^2+(y - 6)^2=(4)(9)$
d $2(x + 2)^2+2(y - 6)^2 = 9$

Explanation:

Step1: Identify the center and radius of circle A

The standard - form of a circle equation is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. For circle A with equation $(x + 2)^2+y^2 = 9$, we can rewrite it as $(x-(-2))^2+(y - 0)^2=3^2$. So the center of circle A is $(-2,0)$ and the radius $r_A=3$.

Step2: Find the center of circle B

Shifting circle A down 6 units means subtracting 6 from the y - coordinate of the center. The x - coordinate remains the same. The center of circle A is $(-2,0)$, so the center of circle B, $(h_B,k_B)=(-2,0 - 6)=(-2,-6)$.

Step3: Find the radius of circle B

The radius of circle B is 2 times the radius of circle A. Since $r_A = 3$, then $r_B=2\times3 = 6$, and $r_B^2=36=(4)(9)$.

Step4: Write the equation of circle B

Using the standard - form of a circle equation $(x - h)^2+(y - k)^2=r^2$ with $(h,k)=(-2,-6)$ and $r^2=(4)(9)$, we get $(x+2)^2+(y + 6)^2=(4)(9)$.

Answer:

A. $(x + 2)^2+(y + 6)^2=(4)(9)$