Question

the common tangent to the circles $x^2+y^2-6x=0$ and $x^2+y^2+2x=0$ forms

• A. Right angle triangle
• B. Right angle isosceles triangle
• C. isosceles triangle
• D. Equilateral triangle

Answer 1

The correct option is D Equilateral triangle

Given circles are
$x^2+y^2-6x=0$
Centre $C_1=(3,0)$ and radius $r_1=3$

$x^2+y^2+2x=0$
Centre $C_2=(-1,0)$ and radius $r_2=1$
$\because C_1{C}_2=r_1+r_2$

Number of common tangents possible are $3.$


$I$ and $E$ divides $C_1$ and $C_2$ in ratio $1:3$ internally and externally.

So,
$I=(\frac{1\left(3\right)+3(-1)}{4},\frac{1\left(0\right)+3\left(0\right)}{4})=(0,0)$

$E=(\frac{1\left(3\right)-3(-1)}{-2},\frac{1\left(0\right)-3\left(0\right)}{-2})=(-3,0)$

For Direct common tangent,
$y-0=m(x+3)$
$\Rightarrow mx-y+3m=0$
Perpendicular distance = Radius
$\Rightarrow \left|\frac{2m}{\sqrt{1+m^2}}\right|=1$
$\Rightarrow 4m^2=m^2+1$
$\Rightarrow m=\pm \frac{1}{\sqrt{3}}$

So, inclination of direct common tangents are ${30}^{\circ }$ and ${150}^{\circ }$


It can be observed that in $△EAB,$
$\angle E=\angle A=\angle B={60}^{\circ }.$
So, it is equilateral triangle.