Q91 / 180 · KCET 2007 · Mathematics

The number of common tangents to the circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}-6 x-8 y-24=0$ is

  • A
    3
  • Correct answer
    B
    4
  • C
    2
  • D
    1

Solution

Given, equation of circles are $$ x^{2}+y^{2}=4 $$ whose radius $=2$ centre $=(0,0)$ and $x^{2}+y^{2}-6 x-8 y-24=0$, whose radius $=\sqrt{9+16-24}=\sqrt{1}=1$ and centre $=(3,4)$ Now, $\quad c_{1} c_{2}=\sqrt{(3-0)^{2}+(4-0)^{2}}$ $$ =\sqrt{9+16}=5 $$ $$ a_{1}+a_{2}=2+1=3 $$ Since, $c_{1} c_{2}>a_{1}+a_{2}$ $\therefore$ Number of common tangents $=4$

Hence, the correct answer is option: 'B'.

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