Problem:
The graphs of x2+y2=4+12x+6y and x2+y2=k+4x+12y intersect when k satisfies a≤k≤b, and for no other values of k. Find b−a.
Answer Choices:
A. 5
B. 68
C. 104
D. 140
E. 144
Solution:
Complete the squares in the two equations to bring them to the form
(x−6)2+(y−3)2=72 and (x−2)2+(y−6)2=k+40
The graphs of these equations are circles. The first circle has radius 7, and the distance between the centers of the circles is 5. In order for the circles to have a point in common, therefore, the radius of the second circle must be at least 2 and at most 12. It follows that 22≤k+40≤122, or −36≤k≤104. Thus b−a=140.