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The number of common tangents to the circles x^2 + y^2 - 4 x - 6 y - 1 2 = 0 and x^2+y^2+6x+18y+26=0, is

JEE Main 2015 — Mathematics Coordinate Geometry

2015mcqeasy

The number of common tangents to the circles x2+y2−4x−6y−12=0{x}^{2} + {y}^{2} - 4 x - 6 y - 1 2 = 0x2+y2−4x−6y−12=0 and x2+y2+6x+18y+26=0{x}^{2}+{y}^{2}+6x+18y+26=0x2+y2+6x+18y+26=0, is

Official previous-year question

Held on 4 Apr 2015 · Verified 6 Jul 2026.

Options

  1. A

    444

  2. B

    111

  3. C

    222

  4. D

    333

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