Circle A in the xy -plane has the equation x + 5 2 + y - 5 2 = 4

Question

Circle A in the xy -plane has the equation x + 5 2 + y - 5 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 + 5 2 + y - 5 2 = k , where k is a constant. What is the value of k ?

Accepted Answer

The correct answer is 16 . An equation of a circle in the xy -plane can be written as x - t 2 + y - u 2 = r 2 , where the center of the circle is t , u , the radius of the circle is r , and where t , u , and r are constants. It&rsquo;s given that the equation of circle A is x + 5 2 + y - 5 2 = 4 , which is equivalent to x + 5 2 + y - 5 2 = 2 2 . Therefore, the center of circle A is - 5 , 5 and the radius of circle A is 2 . It&rsquo;s given that circle B has the same center as circle A and that the radius of circle B is two times the radius of circle A. Therefore, the center of circle B is - 5 , 5 and the radius of circle B is 2 2 , or 4 . Substituting - 5 for t , 5 for u , and 4 for r into the equation x - t 2 + y - u 2 = r 2 yields x + 5 2 + y - 5 2 = 4 2 , which is equivalent to x + 5 2 + y - 5 2 = 16 . It follows that the equation of circle B in the xy -plane is x + 5 2 + y - 5 2 = 16 . Therefore, the value of k is 16 .

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