1. Problem 9 states: Circles P and Q are tangent at S. AB is tangent to both circles at S. Given AB = 16, AP = 12, AQ = 10, find length PQ if it bisects AB. 2. Since AB is tangent at S and PQ bisects AB, point S is midpoint of AB, so AS = SB = $\frac{16}{2} = 8$. 3. Using power of point A relative to circle P: $AP^2 = AS \times AB = 12^2 = 8 \times AB$. Here, $AB$ as length from A to circle P tangent relates to segment AS and radius. 4. But AP given as 12, AQ as 10. Since AB is tangent at S to both circles, the distances AS and SB relate to radii of circles P and Q. 5. We use right triangles: in triangle PAS, $PA = 12$, $AS=8$ so radius of circle P is $r_P = \sqrt{12^2 - 8^2} = \sqrt{144 - 64} = \sqrt{80} = 4\sqrt{5}$. 6. Similarly, for circle Q, triangle QAS has $QA = 10$, $AS=8$, radius of Q is $r_Q = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6$. 7. PQ is the sum of radii because the circles are tangent, so $PQ = r_P + r_Q = 4\sqrt{5} + 6$. --- 8. Problem 10 states: AT tangent to circles K and J at A. ST tangent to circle K at S, RT tangent to circle J at R. Given ST = $2x+7$, RT = $3x+1$, find a. x, b. ST, c. RT, d. AT. 9. Since ST and RT are tangents from T to circles K and J, and AT tangent to both at A, by the tangent-secant theorem, lengths of tangents from T to each circle are equal: $ST = RT$, so $2x + 7 = 3x + 1$ 10. Solve for x: $2x + 7 = 3x + 1 \Rightarrow 7 - 1 = 3x - 2x \Rightarrow 6 = x$ 11. Find ST: $ST = 2(6) + 7 = 12 + 7 = 19$ 12. Find RT: $RT = 3(6) + 1 = 18 + 1 = 19$ 13. Length AT is sum of tangents ST and RT: $AT = ST + RT = 19 + 19 = 38$