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Math  /  Geometry

Question9 Mark for Review
Circle A in the xyx y-plane has the equation (x+5)2+(y−5)2=4(x+5)^{2}+(y-5)^{2}=4. Circle B has the same center as circle AA. The radius of circle BB is two times the radius of circle AA. The equation defining circle B in the xyx y-plane is (x+5)2+(y−5)2=x22(x+5)^{2}+(y-5)^{2}=x_{2}^{2}, where kk is a constant. What is the value of kk ? □\square
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Studdy Solution

STEP 1

1. Circle A has the equation (x+5)2+(y−5)2=4(x+5)^2 + (y-5)^2 = 4.
2. Circle B has the same center as Circle A.
3. The radius of Circle B is twice the radius of Circle A.
4. The equation for Circle B is of the form (x+5)2+(y−5)2=k2(x+5)^2 + (y-5)^2 = k^2.

STEP 2

1. Identify the center and radius of Circle A.
2. Determine the radius of Circle B.
3. Write the equation for Circle B.
4. Solve for the value of kk.

STEP 3

Identify the center and radius of Circle A:
- The equation (x+5)2+(y−5)2=4(x+5)^2 + (y-5)^2 = 4 is in the standard form (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2. - The center (h,k)(h, k) of Circle A is (−5,5)(-5, 5). - The radius rr of Circle A is 4=2\sqrt{4} = 2.

STEP 4

Determine the radius of Circle B:
- The radius of Circle B is twice the radius of Circle A. - Therefore, the radius of Circle B is 2×2=42 \times 2 = 4.

STEP 5

Write the equation for Circle B:
- Since Circle B has the same center as Circle A, its equation is (x+5)2+(y−5)2=(4)2(x+5)^2 + (y-5)^2 = (4)^2.

STEP 6

Solve for the value of kk:
- The equation for Circle B is (x+5)2+(y−5)2=16(x+5)^2 + (y-5)^2 = 16. - Thus, k2=16k^2 = 16. - Therefore, k=16=4k = \sqrt{16} = 4.
The value of kk is:
4 \boxed{4}

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