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. $\sqrt{17}units$
• B. $3units$
• C. $9units$
• D. $\sqrt{15}units$

Answer 1

The correct option is B $3units$

We have been given two equation for two circles:

Circle1 $=x^2+y^2-2x+4y+4=0$

Centerl $=(1,-2)$; radius $=1=r_1$

$\mathrm{circle}2=x^2+y^2+4x-2y+1=0$

Center $2=(-2,1)$; radius $=2=r_2$

The formula for the distance between centers of the lireles:

$d=($ Center 1) $\left(\mathrm{Center}2\right)=\sqrt{18}=3\sqrt{2}$

Now; $d>r_1+r_2$ [ no intersection ]

Length of transverse common tangent; $L=\sqrt{d^2+{(r_1+r_2)}^2}$

$=\sqrt{(3\sqrt{2}{)}^2+(1+2{)}^2}=\sqrt{9}$

$=3\text{units}$