# 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 $b-a$.

$\mathrm{(A) \ }5 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \ }104 \qquad \mathrm{(D) \ }140 \qquad \mathrm{(E) \ }144$

## Solution

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 if $2\leq r \leq 12$.

From $\sqrt{40+k} \geq 2$ we get that $k\geq -36$. From $\sqrt{40+k}\leq 12$ we get $k\leq 104$.

Therefore $b-a = 104 - (-36) = \boxed{140}$.