1998 AHSME Problems/Problem 23

Problem

The graphs of $x^2 + y^2 = 4 + 12x + 6y$ and $x^2 + y^2 = k + 4x + 12y$ intersect when $k$ satisfies $a \le k \le b$ , and for no other values of $k$ . Find $b-a$ .

Both sets of points are quite obviously circles. To show this, we can rewrite each of them in the form $(x-x_0)^2 + (y-y_0)^2 = r^2$ .

The first curve becomes $(x-6)^2 + (y-3)^2 = 7^2$ , which is a circle centered at $(6,3)$ with radius $7$ .

The second curve becomes $(x-2)^2 + (y-6)^2 = 40+k$ , which is a circle centered at $(2,6)$ with radius $r=\sqrt{40+k}$ .

The distance between the two centers is $5$ , and therefore the two circles intersect iff $2\leq r \leq 12$ .