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 centers of the three circles be $C_1, C_2, C_3$ with radii $a, b, c$ respectively. Since the x-axis is a common tangent, the y-coordinates of the centers are $a, b, c$. Let the x-coordinates of the centers be $x_a, x_b, x_c$. The distance between the centers of two externally touching circles with radii $r_i$ and $r_j$ is $r_i + r_j$. The distance between centers $(x_i, r_i)$ and $(x_j, r_j)$ is $\sqrt{(x_i-x_j)^2 + (r_i-r_j)^2}$. Thus, $(x_i-x_j)^2 + (r_i-r_j)^2 = (r_i+r_j)^2$, which simplifies to $|x_i-x_j| = 2\sqrt{r_i r_j}$.

If the circle with radius $b$ is between the circles with radii $a$ and $c$ (in terms of x-coordinates), then $x_b - x_a = 2\sqrt{ab}$ and $x_c - x_b = 2\sqrt{bc}$. Summing these gives $x_c - x_a = 2\sqrt{ab} + 2\sqrt{bc}$. We also know $x_c - x_a = 2\sqrt{ac}$. Therefore, $2\sqrt{ac} = 2\sqrt{ab} + 2\sqrt{bc}$. Dividing by $2\sqrt{abc}$ yields $\frac{1}{\sqrt{b}} = \frac{1}{\sqrt{c}} + \frac{1}{\sqrt{a}}$. This is option (1).

If the circle with radius $a$ is between the circles with radii $b$ and $c$, then $x_a - x_b = 2\sqrt{ab}$ and $x_c - x_a = 2\sqrt{ac}$. Summing these gives $x_c - x_b = 2\sqrt{ab} + 2\sqrt{ac}$. We also know $x_c - x_b = 2\sqrt{bc}$. Therefore, $2\sqrt{bc} = 2\sqrt{ab} + 2\sqrt{ac}$. Dividing by $2\sqrt{abc}$ yields $\frac{1}{\sqrt{c}} = \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}}$.

If the circle with radius $c$ is between the circles with radii $a$ and $b$, then $x_c - x_a = 2\sqrt{ac}$ and $x_b - x_c = 2\sqrt{bc}$. Summing these gives $x_b - x_a = 2\sqrt{ac} + 2\sqrt{bc}$. We also know $x_b - x_a = 2\sqrt{ab}$. Therefore, $2\sqrt{ab} = 2\sqrt{ac} + 2\sqrt{bc}$. Dividing by $2\sqrt{abc}$ yields $\frac{1}{\sqrt{c}} = \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}}$.

Given $a < b < c$, we have $\frac{1}{\sqrt{a}} > \frac{1}{\sqrt{b}} > \frac{1}{\sqrt{c}}$. Option (1): $\frac{1}{\sqrt{b}} = \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{c}}$. This is impossible since $\frac{1}{\sqrt{a}} > \frac{1}{\sqrt{b}}$ and $\frac{1}{\sqrt{c}} > 0$. Option (4): $\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}}$. This is possible if the circle with radius $a$ is positioned between the other two. This corresponds to the case where the middle circle in x-coordinate has radius $a$. However, the derivation shows that if the circle with radius $b$ is in the middle, we get $\frac{1}{\sqrt{b}} = \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{c}}$. This is incorrect.

Let's re-evaluate the x-coordinate ordering. Case 1: Circle $b$ is between $a$ and $c$. $x_b-x_a = 2\sqrt{ab}$, $x_c-x_b = 2\sqrt{bc}$. Then $x_c-x_a = 2\sqrt{ab}+2\sqrt{bc}$. Also $x_c-x_a = 2\sqrt{ac}$. So $2\sqrt{ac} = 2\sqrt{ab}+2\sqrt{bc} \implies \frac{1}{\sqrt{b}} = \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{c}}$. This contradicts $a<b<c$ as $\frac{1}{\sqrt{a}} > \frac{1}{\sqrt{b}}$.

Case 2: Circle $a$ is between $b$ and $c$. $x_a-x_b = 2\sqrt{ab}$, $x_c-x_a = 2\sqrt{ac}$. Then $x_c-x_b = 2\sqrt{ab}+2\sqrt{ac}$. Also $x_c-x_b = 2\sqrt{bc}$. So $2\sqrt{bc} = 2\sqrt{ab}+2\sqrt{ac} \implies \frac{1}{\sqrt{c}} = \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{a}}$. This is consistent with $a<b<c$.

Case 3: Circle $c$ is between $a$ and $b$. $x_c-x_a = 2\sqrt{ac}$, $x_b-x_c = 2\sqrt{bc}$. Then $x_b-x_a = 2\sqrt{ac}+2\sqrt{bc}$. Also $x_b-x_a = 2\sqrt{ab}$. So $2\sqrt{ab} = 2\sqrt{ac}+2\sqrt{bc} \implies \frac{1}{\sqrt{c}} = \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}}$. This is also consistent.

The question implies a single correct statement. The standard result for three mutually touching circles with a common tangent is $\frac{1}{\sqrt{r_1}} = \frac{1}{\sqrt{r_2}} + \frac{1}{\sqrt{r_3}}$ where $r_2$ is the middle radius. However, the options do not reflect this directly.

Let's re-examine the case where the circle with radius 'b' is in the middle. The condition is $x_b - x_a = 2\sqrt{ab}$ and $x_c - x_b = 2\sqrt{bc}$. This leads to $x_c - x_a = 2\sqrt{ab} + 2\sqrt{bc}$. Since $x_c - x_a = 2\sqrt{ac}$, we have $2\sqrt{ac} = 2\sqrt{ab} + 2\sqrt{bc}$. Dividing by $2\sqrt{abc}$, we get $\frac{1}{\sqrt{b}} = \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{c}}$. This is option (1). However, since $a<b<c$, we have $\frac{1}{\sqrt{a}} > \frac{1}{\sqrt{b}} > \frac{1}{\sqrt{c}}$. Thus $\frac{1}{\sqrt{a}} + \frac{1}{\sqrt{c}} > \frac{1}{\sqrt{b}}$, so option (1) is impossible.

Consider the case where the circle with radius 'a' is in the middle. Then $x_a - x_b = 2\sqrt{ab}$ and $x_c - x_a = 2\sqrt{ac}$. This implies $x_c - x_b = 2\sqrt{ab} + 2\sqrt{ac}$. Since $x_c - x_b = 2\sqrt{bc}$, we have $2\sqrt{bc} = 2\sqrt{ab} + 2\sqrt{ac}$. Dividing by $2\sqrt{abc}$, we get $\frac{1}{\sqrt{c}} = \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{a}}$. This is option (4) with $a$ and $c$ swapped. So option (4) is $\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}}$. This is not derivable from this configuration.

Let's use Descartes' Theorem. For four mutually tangent circles with curvatures $k_i = \pm 1/r_i$, the theorem states $(k_1+k_2+k_3+k_4)^2 = 2(k_1^2+k_2^2+k_3^2+k_4^2)$. Here, we have three circles tangent to each other and to the x-axis. The x-axis can be considered a circle of infinite radius, hence curvature $k_4 = 0$. The curvatures of the three circles are $k_a = 1/a$, $k_b = 1/b$, $k_c = 1/c$ (since they are externally tangent to each other and the x-axis). So, $(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2 = 2(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2})$. This does not seem to lead to the options easily.

Let's go back to the distance between centers. The condition $|x_i - x_j| = 2\sqrt{r_i r_j}$ is correct. Let the centers be $C_a = (x_a, a)$, $C_b = (x_b, b)$, $C_c = (x_c, c)$. Assume $x_a < x_b < x_c$. Then $x_b - x_a = 2\sqrt{ab}$ and $x_c - x_b = 2\sqrt{bc}$. Adding these, $x_c - x_a = 2\sqrt{ab} + 2\sqrt{bc}$. We also know $x_c - x_a = 2\sqrt{ac}$. 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, since $a < b < c$, we have $\frac{1}{\sqrt{a}} > \frac{1}{\sqrt{b}} > \frac{1}{\sqrt{c}}$. So $\frac{1}{\sqrt{a}} + \frac{1}{\sqrt{c}} > \frac{1}{\sqrt{b}}$. Thus, this arrangement is impossible.

Let's assume the circles are ordered such that the middle circle in terms of x-coordinate has radius 'a'. So, $x_b < x_a < x_c$. Then $x_a - x_b = 2\sqrt{ab}$ and $x_c - x_a = 2\sqrt{ac}$. Adding these, $x_c - x_b = 2\sqrt{ab} + 2\sqrt{ac}$. We also know $x_c - x_b = 2\sqrt{bc}$. So, $2\sqrt{bc} = 2\sqrt{ab} + 2\sqrt{ac}$. Dividing by $2\sqrt{abc}$: $\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{c}} + \frac{1}{\sqrt{b}}$. This is option (4). This is consistent with $a < b < c$, as $\frac{1}{\sqrt{a}}$ is the largest.

Let's assume the circles are ordered such that the middle circle in terms of x-coordinate has radius 'c'. So, $x_a < x_c < x_b$. Then $x_c - x_a = 2\sqrt{ac}$ and $x_b - x_c = 2\sqrt{bc}$. Adding these, $x_b - x_a = 2\sqrt{ac} + 2\sqrt{bc}$. We also know $x_b - x_a = 2\sqrt{ab}$. So, $2\sqrt{ab} = 2\sqrt{ac} + 2\sqrt{bc}$. Dividing by $2\sqrt{abc}$: $\frac{1}{\sqrt{c}} = \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{a}}$.

The question implies a single correct statement. Given $a < b < c$, the relation $\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}}$ (Option 4) is the one that can be true. This happens when the circle with the smallest radius is positioned between the other two in terms of x-coordinates.

Final Check: If $\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}}$, then $\frac{1}{\sqrt{a}}$ is the largest, which is consistent with $a$ being the smallest radius. This implies that the arrangement of circles such that the circle with radius $a$ is in the middle (horizontally) is the correct one. The condition for this arrangement is $x_c - x_b = 2\sqrt{bc}$ and $x_a - x_b = 2\sqrt{ab}$ and $x_c - x_a = 2\sqrt{ac}$. This leads to $x_c - x_b = (x_c - x_a) + (x_a - x_b) = 2\sqrt{ac} + 2\sqrt{ab}$. So $2\sqrt{bc} = 2\sqrt{ac} + 2\sqrt{ab}$. Dividing by $2\sqrt{abc}$, we get $\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{c}} + \frac{1}{\sqrt{b}}$. This is option (4).

The other options are incorrect: (2) a, b, c are in A.P. means $2b = a+c$. This is not generally true. (3) $\sqrt{a}$, $\sqrt{b}$, $\sqrt{c}$ are in A.P. means $2\sqrt{b} = \sqrt{a} + \sqrt{c}$. This is also not generally true.

Therefore, option (4) is the correct statement.