Question

Let $S_1$ and $S_2$ be two fixed circles touching each other externally with radius 2 and 3 respectively. Let $S_3$ be a variable circle touching internally both $S_1$ and $S_2$ at points A and B respectively. The tangents to $S_3$ at A and B meet at T, and TA = 4.

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (D)

8

Answer 2

Let $C_1, C_2, C_3$ be the centers and $r_1, r_2, r_3$ be the radii of circles $S_1, S_2, S_3$ respectively. We are given $r_1 = 2$ and $r_2 = 3$. Since $S_1$ and $S_2$ touch externally, the distance between their centers is $C_1C_2 = r_1 + r_2 = 2 + 3 = 5$. Since $S_3$ touches $S_1$ internally, $C_3C_1 = |r_3 - r_1| = |r_3 - 2|$. Assuming $r_3 > r_1$, $C_3C_1 = r_3 - 2$. Since $S_3$ touches $S_2$ internally, $C_3C_2 = |r_3 - r_2| = |r_3 - 3|$. Assuming $r_3 > r_2$, $C_3C_2 = r_3 - 3$. For these distances to form a triangle, we must have $C_3C_1 + C_3C_2 > C_1C_2$, which implies $(r_3-2) + (r_3-3) > 5$, so $2r_3 - 5 > 5$, leading to $2r_3 > 10$, or $r_3 > 5$. This eliminates options (A) 2 and (B) 4.

The tangents to $S_3$ at A and B meet at T, and TA = 4. This implies $TA = TB = 4$. Also, $C_3A \perp TA$ and $C_3B \perp TB$, and $C_3A = C_3B = r_3$. Consider the right-angled triangle $\triangle C_3AT$. We have $C_3T^2 = C_3A^2 + TA^2 = r_3^2 + 4^2 = r_3^2 + 16$. Let $\angle AC_3T = \phi$. Then $\tan \phi = \frac{TA}{C_3A} = \frac{4}{r_3}$. The angle $\angle AC_3B = 2\phi$. In $\triangle C_1C_2C_3$, by the Law of Cosines: $C_1C_2^2 = C_3C_1^2 + C_3C_2^2 - 2(C_3C_1)(C_3C_2)\cos(\angle C_1C_3C_2)$. We have $\angle C_1C_3C_2 = \angle AC_3B = 2\phi$. Using $\cos(2\phi) = \frac{1-\tan^2\phi}{1+\tan^2\phi} = \frac{1-(4/r_3)^2}{1+(4/r_3)^2} = \frac{r_3^2-16}{r_3^2+16}$. Substituting into the Law of Cosines: $5^2 = (r_3-2)^2 + (r_3-3)^2 - 2(r_3-2)(r_3-3)\left(\frac{r_3^2-16}{r_3^2+16}\right)$.

Testing $r_3=8$: $C_3C_1 = 8-2=6$, $C_3C_2 = 8-3=5$. $\cos(2\phi) = \frac{8^2-16}{8^2+16} = \frac{64-16}{64+16} = \frac{48}{80} = \frac{3}{5}$. LHS = $25$. RHS = $6^2 + 5^2 - 2(6)(5)(\frac{3}{5}) = 36+25 - 60(\frac{3}{5}) = 61 - 36 = 25$. LHS = RHS, so $r_3=8$ is correct.