Question

Three circles of radii a, b, c(a < b < c) touch each other externally. If they have x-axis as a common tangent, then :

• A. $\frac{1}{\sqrt{b}} = \frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}$
• B. a, b, c are in A. P.
• C. $\sqrt{a}, \sqrt{b}, \sqrt{c}$ are in A. P.
• D. $\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$

Answer 1

Answer: (D)

$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$

Answer 2

Let the radii of the three circles be $r_1, r_2, r_3$. Since the x-axis is a common tangent, the centers of the circles are $C_i = (x_i, r_i)$. When two circles with radii $r_i$ and $r_j$ touch externally, the distance between their centers is $r_i+r_j$. The distance between centers $C_i=(x_i, r_i)$ and $C_j=(x_j, r_j)$ is $\sqrt{(x_j-x_i)^2 + (r_j-r_i)^2}$. So, $(x_j-x_i)^2 + (r_j-r_i)^2 = (r_i+r_j)^2$. This gives $(x_j-x_i)^2 = (r_i+r_j)^2 - (r_j-r_i)^2 = 4r_ir_j$. Thus, the horizontal distance between the centers of two tangent circles is $|x_j-x_i| = 2\sqrt{r_i r_j}$.

For the three circles with radii $a, b, c$, the horizontal distances between their centers are $2\sqrt{ab}$, $2\sqrt{bc}$, and $2\sqrt{ac}$. Since the circles touch each other externally and have a common tangent, their centers must be collinear in terms of their x-coordinates. This means one of these distances must be the sum of the other two.

Given $a < b < c$, we have $\sqrt{a} < \sqrt{b} < \sqrt{c}$. This implies $\sqrt{ab} < \sqrt{ac}$ and $\sqrt{bc} < \sqrt{ac}$. Also $\sqrt{ab} < \sqrt{bc}$. The order of distances is $2\sqrt{ab} < 2\sqrt{bc} < 2\sqrt{ac}$.

The largest distance is $2\sqrt{ac}$, which is between the centers of circles with radii $a$ and $c$. This distance must be the sum of the other two distances: $2\sqrt{ab}$ and $2\sqrt{bc}$. This configuration implies that the circle with radius $b$ is positioned between the circles with radii $a$ and $c$ along the x-axis. So, $2\sqrt{ac} = 2\sqrt{ab} + 2\sqrt{bc}$. Dividing by $2\sqrt{abc}$: $\frac{1}{\sqrt{b}} = \frac{1}{\sqrt{c}} + \frac{1}{\sqrt{a}}$. This is option (1).

However, the relation $\frac{1}{\sqrt{r_i}} + \frac{1}{\sqrt{r_k}} = \frac{1}{\sqrt{r_j}}$ holds when $r_j$ is the radius of the circle whose center is between the centers of the other two. The distance between centers of circles with radii $r_i$ and $r_k$ is $2\sqrt{r_i r_k}$. The distances are $2\sqrt{ab}$, $2\sqrt{bc}$, $2\sqrt{ac}$. Since $a < b < c$, we have $\sqrt{ab} < \sqrt{bc} < \sqrt{ac}$. The largest distance is $2\sqrt{ac}$. The sum of the two smaller distances is $2\sqrt{ab} + 2\sqrt{bc}$. If the circle with radius $a$ is in the middle, then $2\sqrt{bc} = 2\sqrt{ab} + 2\sqrt{ac}$. Dividing by $2\sqrt{abc}$ gives $\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{c}} + \frac{1}{\sqrt{b}}$. If the circle with radius $b$ is in the middle, then $2\sqrt{ac} = 2\sqrt{ab} + 2\sqrt{bc}$. Dividing by $2\sqrt{abc}$ gives $\frac{1}{\sqrt{b}} = \frac{1}{\sqrt{c}} + \frac{1}{\sqrt{a}}$. If the circle with radius $c$ is in the middle, then $2\sqrt{ab} = 2\sqrt{bc} + 2\sqrt{ac}$, which is impossible since $a,b,c > 0$.

The condition $a<b<c$ implies that $1/\sqrt{a} > 1/\sqrt{b} > 1/\sqrt{c}$. Let $X=1/\sqrt{a}, Y=1/\sqrt{b}, Z=1/\sqrt{c}$. We have $X > Y > Z$. The relation $\frac{1}{\sqrt{r_i}} + \frac{1}{\sqrt{r_k}} = \frac{1}{\sqrt{r_j}}$ implies that $1/\sqrt{r_j}$ is the largest of the three reciprocals of square roots, meaning $r_j$ is the smallest radius. Since $a$ is the smallest radius ($a<b<c$), it must be $r_j$. Therefore, the relation is $\frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}} = \frac{1}{\sqrt{a}}$. This is option (4). This implies the circle with radius $a$ is positioned between the circles with radii $b$ and $c$ in terms of their x-coordinates. The distances are $2\sqrt{ab}, 2\sqrt{ac}, 2\sqrt{bc}$. Since $a<b<c$, we have $\sqrt{ab} < \sqrt{ac} < \sqrt{bc}$. The largest distance is $2\sqrt{bc}$. This must be the sum of the other two: $2\sqrt{bc} = 2\sqrt{ab} + 2\sqrt{ac}$. Dividing by $2\sqrt{abc}$ gives $\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{c}} + \frac{1}{\sqrt{b}}$.