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 $=2$