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JEE Main 201226 MayOnline
Exam
JEE Main
Year
2012
Date
26 May
Mode
Online
Subject
Mathematics
Chapter
Circle

JEE Main 2012 · 26 May · Online · Mathematics · Circle

The number of common tangents of the circles given by x^2+y^2-8 x-2...

The number of common tangents of the circles given by x2+y2−8x−2y+1=0x^2+y^2-8 x-2 y+1=0 and x2+y2+6x+8y=0x^2+y^2+6 x+8 y=0 is

Options

  1. Aone
  2. Bfour
  3. Ctwo
  4. Dthree

Correct answer

C. two

Step-by-step solution

Given circles are x2+y2−8x−2y+1=0x^2+y^2-8 x-2 y+1=0 and x2+y2+6x+8y=0x^2+y^2+6 x+8 y=0 Their centres and radius are C_1(4,1), r_1=16=\sqrt{16}=4 C_2(-3,-4), r_2=25=\sqrt{25}=5 Now, C1C2=49+25=74C_1 C_2=\sqrt{49+25}=\sqrt{74} r_1-r_2=-1, r_1+r_2=9 Since, r1−r2<C1C2<r1+r2∴r_1-r_2 \lt C_1 C_2 \lt r_1+r_2\therefore Number of common tangents =2=2
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