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

<math> \mathrm{(A) \ }5 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \ }104 \qquad \mathrm{(D) \ }140 \qquad \mathrm{(E) \ }144 </math>

== Solution ==

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

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

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

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

From <math>\sqrt{40+k} \geq 2</math> we get that <math>k\geq -36</math>. From <math>\sqrt{40+k}\leq 12</math> we get <math>k\leq 104</math>. 

Therefore <math>b-a = 104 - (-36) = \boxed{140}</math>.

== See also ==
{{AHSME box|year=1998|num-b=22|num-a=24}}