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
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
one
four
two
three
Solution
Given circles are $x^2+y^2-8 x-2 y+1=0$ and $x^2+y^2+6 x+8 y=0$ Their centres and radius are
$
C_1(4,1), r_1=\sqrt{16}=4
$
$
C_2(-3,-4), 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 $=2$