Question

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

• A. one
• B. four
• C. two
• D. three

Answer 1

Answer: (C)

Given circles are
$x^{2}+y^{2}-8x-2y+1=0$
and $x^{2} +y^{2} + 6x + 8y = 0$
Their centres and radius are
$C_{1}\left(4, 1\right), r_{1} = \sqrt{16} = 4$
$C_{2}\left(-3,-4\right),r_{2}= \sqrt{25}=5$
Now, $C_{1}C_{2} = \sqrt{49+25} = \sqrt{74}$
$r_{1}-r_{2}=-1, r_{1}+r_{2} = 9$
Since, $r_{1} -r_{2} < C_{1}C_{2} < r_{1}+r_{2}$
$\therefore $ Number of common tangents $= 2$