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

JEE Main 2012 — Mathematics Coordinate Geometry

2012mcqmedium

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

Official previous-year question

Held on 26 May 2012 · Verified 6 Jul 2026.

Options

  1. A

    one

  2. B

    four

  3. C

    two

  4. D

    three

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