The length of the transverse common tangent of the circles $x^2+y^2-2 x+4 y+4=0$ and $x^2+y^2+4 x-2 y+1=0$ is
The length of the transverse common tangent of the circles $x^2+y^2-2 x+4 y+4=0$ and $x^2+y^2+4 x-2 y+1=0$ is
$\sqrt{3}$
$\sqrt{17}$
$\sqrt{15}$
3
Solution
$S_1: x^2+y^2-2 x+4 y+4=0$
Centre, $\quad C_1(1,-2)$ and $r_1=1$
and $S_2: x^2+y^2+4 x-2 y+1=0$
Centre $\quad C_2(-2,1)$ and $r_2=2$
Distance between centres, $d$ is
$
\begin{gathered}
d=\sqrt{(1+2)^2+(-2-1)^2} \\
d=\sqrt{18}=3 \sqrt{2} \\
d>r_1+r_2
\end{gathered}
$
$\therefore S_1$ and $S_2$ are not intersecting each other.
The length of transversal common tangent is
$
\begin{aligned}
& L=\sqrt{d^2-\left(r_1+r_2\right)^2}=\sqrt{(3 \sqrt{2})^2-9}=\sqrt{9} \\
& L=3 \text { units }
\end{aligned}
$