If the circles $x^2+y^2-2 x+4 y+c=0$ and $x^2+y^2+2 x-$ $4 y+c=0$ have four common tangents, then
If the circles $x^2+y^2-2 x+4 y+c=0$ and $x^2+y^2+2 x-$ $4 y+c=0$ have four common tangents, then
$\mathrm{c} < 0$
$-2 < $ c $ < 2$
$0 < $ c $ < 5$
$\mathrm{c}>0$
Solution
The given equation of circle is as follows :-
$x^2+y^2-2 x+4 y+c=0$ ...(i)
$x^2+y^2+2 x-4 y+c=0$ ...(ii)
Radius and centre of circle (i) is
$r_1=\sqrt{(-1)^2+(2)^2-c}=\sqrt{5-c}$ and $c_1=(+1,-2)$
Radius and centre of circle (ii) is
$r_2=\sqrt{(1)^2+(-2)^2-c}=\sqrt{5-c}$ and $c_2=(-1,2)$
$\because$ Circle (i) and (ii) have four common tangents.
$\begin{aligned} & \therefore \quad r_1+r_2 < \left|c_1 c_2\right| \\ & \Rightarrow \quad \sqrt{5-c}+\sqrt{5-c} < \sqrt{4+16} \\ & \Rightarrow \quad 2 \sqrt{5-c} < \sqrt{20} \Rightarrow 4(5-c) < 20\end{aligned}$
$\Rightarrow 5-c < 5 \Rightarrow c>0$ ...(i)
Radius of circles should be positive.
$\begin{aligned} & \therefore r_1>0 \\ & \Rightarrow \sqrt{5-c}>0 \Rightarrow 5-c>0\end{aligned}$
$\Rightarrow \quad c < 5$ ...(ii)
From eqn. (i) and (ii) $0 < c < 5$
Option (3) is correct.