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

(A) $\frac{1}{\sqrt{\mathrm{a}}}=\frac{1}{\sqrt{\mathrm{b}}}+\frac{1}{\sqrt{\mathrm{c}}}$

(B) $\frac{1}{\sqrt{\mathrm{b}}}=\frac{1}{\sqrt{\mathrm{a}}}+\frac{1}{\sqrt{\mathrm{c}}}$

(D) $\sqrt{a}, \sqrt{b}, \sqrt{c}$ are in A.P.

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Circle

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) $\frac{1}{a} = \frac{1}{b} + \frac{1}{c}$ ✓ Correct
(B) $\frac{1}{b} = \frac{1}{a} + \frac{1}{c}$
(C) $a, b, c$ are in A.P.
(D) $a, b, c$ are in A.P.

Answer: (A)

Solution

Each circle touches the x-axis, so its centre lies at a height equal to its radius. For two externally tangent circles sharing the x-axis as common tangent, the horizontal distance between their points of contact is

2 r_{1} r_{2}

. Lining the three up, the contact distances satisfy

2 b c = 2 ab + 2 a c

, and dividing through by

2 ab c

gives

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

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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