Circle A in the xy-plane has the equation (x+3)^2+(y-3)^2=4. Circle B has the same center as circle A. The radius of circle B is two times the radius of Circle A. The equation defining circle B in the xy-plane is (x+3)^2+(y-3)^2=k, where k is a constant. What is the value of k?

Circle A in the xy-plane has the equation (x+3)^2+(y-3)^2=4. Circle B has the same center... | SAT Circles Practice

Explanation:

1. Determine the center and radius of Circle A:

The equation of Circle A is given as $(x+3)^2+(y-3)^2=4$ .

This is in the standard form $(x-h)^2+(y-k)^2=r^2$ , where $(h,k)$ is the center and $r$ is the radius.

For Circle A, the center is $(-3, 3)$ and $r\_A^2 = 4$ , so the radius $r\_A = \sqrt{4} = 2$ .

2. Determine the radius of Circle B:

The radius of Circle B ( $r\_B$ ) is two times the radius of Circle A.

$r\_B = 2 \times r\_A = 2 \times 2 = 4$ .

3. Determine the equation of Circle B and the value of k:

Circle B has the same center as Circle A, which is $(-3, 3)$ .

The equation of Circle B is $(x - (-3))^2 + (y - 3)^2 = r\_B^2$

$(x+3)^2 + (y-3)^2 = 4^2$

$(x+3)^2 + (y-3)^2 = 16$ .

The problem states the equation for Circle B is $(x+3)^2+(y-3)^2=k$ .

Comparing the two forms for Circle B, we see that $k = 16$ .

Conclusion:

The value of k is $16$ .