Question

Given a line segment AB, A = (0,0) and B(a,0). Three circles $S_1, S_2, S_3$ of radius R are centred at the end points and the midpoint of the line segment AB. A fourth circle $S_4$ is drawn touching the 3 given circles.

If 0 < R < $\frac{a}{4}$, then number of possible circle $S_4$ is:

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (D)

8

Answer 2

The problem involves finding the number of circles tangent to three given circles, which is a classic instance of the Problem of Apollonius. The three circles $S_1, S_2, S_3$ have centers at $(0,0)$, $(a,0)$, and $(a/2,0)$ respectively, and all share the same radius $R$. The condition $0 < R < \frac{a}{4}$ is crucial because it ensures that the three circles are disjoint and none is contained within another.

For three circles in such a configuration (disjoint and none inside another), the Problem of Apollonius guarantees that there are exactly 8 distinct circles tangent to all three. These solutions arise from the $2^3=8$ possible combinations of internal and external tangencies between the fourth circle $S_4$ and the given three circles $S_1, S_2, S_3$.

The specific arrangement of the centers on a line does not reduce the number of solutions from 8 in this scenario. The symmetry of the setup about the x-axis means that solutions typically come in pairs $(x,y)$ and $(x,-y)$ for the center of $S_4$, unless the center lies on the x-axis. It can be shown that for this specific problem with the given condition, no solution's center lies on the x-axis. Therefore, there are 4 solutions with centers above the x-axis and 4 solutions with centers below the x-axis, totaling 8 possible circles $S_4$.