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JEE Main 20199 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2019 · 9 January · Morning Shift · Mathematics · Circle

Three circles of radii a, b, c, (a < b < c) touch each other...

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

Options

  1. A1a=1b+1c\dfrac{1}{\sqrt{a}} = \dfrac{1}{\sqrt{b}} + \dfrac{1}{\sqrt{c}}
  2. Ba,b,ca, b, c are in A.P.
  3. C1b=1a+1c\dfrac{1}{\sqrt{b}} = \dfrac{1}{\sqrt{a}} + \dfrac{1}{\sqrt{c}}
  4. Da,b,c\sqrt{a}, \sqrt{b}, \sqrt{c} are in A.P.

Correct answer

A. 1a=1b+1c\dfrac{1}{\sqrt{a}} = \dfrac{1}{\sqrt{b}} + \dfrac{1}{\sqrt{c}}

Step-by-step solution

We have,

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

LM=PN=(a+b)2−(b−a)2=2abLM = PN = \sqrt{{\left(a + b\right)}^{2} - {\left(b - a\right)}^{2}} = 2\sqrt{ab}

NQ=MS=(a+c)2−(c−a)2=2acNQ = MS = \sqrt{{\left(a + c\right)}^{2} - {\left(c - a\right)}^{2}} = 2\sqrt{ac}

PQ=ST=(b+c)2−(c−b)2=2bcPQ = ST = \sqrt{{\left(b + c\right)}^{2} - {\left(c - b\right)}^{2}} = 2\sqrt{bc}

∵∵ PQ=PN+NQPQ = PN + NQ

⇒2bc=2ab+2ac⇒ 2\sqrt{bc} = 2\sqrt{ab} + 2\sqrt{ac}

⇒1a=1b+1c⇒ \dfrac{1}{\sqrt{a}} = \dfrac{1}{\sqrt{b}} + \dfrac{1}{\sqrt{c}}

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

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