Question

The length of the transverse common tangent of the circles $x^2+y^2-2x+4y+4=0$ and $x^2+y^2+4x-2y+1=0$ is

• A. $\sqrt{3}$
• B. $\sqrt{17}$
• C. $\sqrt{15}$
• D. 3

Answer 1

Answer: (D)

$S_1:x^2+y^2-2x+4y+4=0$
Centre, $C_1(1,-2)$ and $r_1=1$
and $S_2:x^2+y^2+4x-2y+1=0$
Centre $C_2(-2,1)$ and $r_2=2$
Distance between centres, $d$ is
$d=\sqrt{(1+2{)}^2+(-2-1{)}^2}$
$d=\sqrt{18}=3\sqrt{2}$
$\because d>r_1+r_2$
$\therefore S_1$ and $S_2$ are not intersecting each other.
The length of transversal common tangent is
$L=\sqrt{d^2-{\left(r_1+r_2\right)}^2}=\sqrt{(3\sqrt{2}{)}^2-9}=\sqrt{9}$
$L=3$ units