Circles
Circle
Allen Star Batch
Grade 11
Question:
Consider a series of 'n' concentric circles $C_1, C_2, C_3, ..., C_n$ with radii $r_1, r_2, r_3, ..., r_n$ respectively, such that $n > r_2, ..., > r_n$ and $\sum r_i = 20$. If the tangents drawn from any point on $C_{i+1}$ are such that the chord of contact is a tangent to $C_{i+2}$ $(i = 1, 2, 3, ...)$ and the angle between the tangents from any point on $C_1$ to $C_2$ is $\frac{\pi}{3}$, then find the values of $\lim_{n \to \infty} \sum_{i=1}^{n} r_i$.
Step-by-Step Solution
Key Concept: Recognize that concentric circles with tangent and chord-of-contact conditions form a geometric progression with ratio determined by the angle between tangents.
Select the center of the circle series as origin $O$. Tangents from a point $P(a\cos\theta, a\sin\theta)$ on circle $S_1: x^2+y^2=a^2$ touch circle $S_2: x^2+y^2=b^2$ such that the chord of contact touches $S_3: x^2+y^2=c^2$. This implies $b^2 = ac$, so $a, b, c$ are in geometric progression. Since $\angle OPQ = \frac{\pi}{6}$ gives $\sin\frac{\pi}{6} = \frac{OA}{OP} = \frac{b}{a} = \frac{1}{2}$, the common ratio is $r = \frac{1}{2}$. Therefore $\lim_{n\to\infty}\sum_{i=1}^{n} r_i = \frac{r_1}{1-\frac{1}{2}} = 2r_1 = 2(20) = 40$.
Correct Answer: 40