Question

Area of triangle formed by common tangents to the circle $x^2+y^2-6x=0and{x}^2+y^2+2x=0$ is

• A. $3\sqrt{3}$
• B. $2\sqrt{3}$
• C. $9\sqrt{3}$
• D. $6\sqrt{3}$

Answer 1

The correct option is A $3\sqrt{3}$

In $S_2$

$x^2+y^2-6x=0$

$x^2+y^2-6x+9=9$

${(x-3)}^2+y^2=9$

$Centre(3,0)$

$Radius=3$

In $S_1$

$x^2+y^2+2x=0$

$x^2+y^2+2x+1=1$

${(x+1)}^2+y^2=1$

$Centre(-1,0)$

$Radius=1$

$c_1{c}_2=r_1+r_2$

$\Rightarrow 2$ direct common tangents 1 transverse commom tangent.

Common tangent$=S_1-S_2$

$=(x^2+y^2+2x)-(x^2+y^2-6x)=8x=0$

$\Rightarrow x=0$ is one transverse common tangent.

Direct common tangents meet at a point where centres$(-1,0),(3,0)$ are divided in the ratio $1:3$ externally.

$\Rightarrow x=\frac{(-1)\times 3-\left(3\right)\times 1}{3-1}=-3$

$y=0$

Direct common tangent intersect at $(-3,0)$

let equation be $(y-0)=mx+3$

$y=mx+3$ touch $x^2+y^2+2x=0$

$\Rightarrow radius=1=$ perpendicular distance from $(1,0)$ to $y=mx+3$

$=\left|\frac{-m+3m}{\sqrt{1+m^2}}\right|$

$1+m^2={4m}^2$

${3m}^2=1$

$\therefore m=\pm \frac{1}{\sqrt{3}}$

$\therefore$ direct common tangents are

$y=\frac{x+3}{\sqrt{3}},y=\frac{-(x+3)}{\sqrt{3}}$

$y=\frac{x+3}{\sqrt{3}}$ intersect $x=0$ at $y=\sqrt{3}$

$y=\frac{-(x+3)}{\sqrt{3}}$ intersect $x=0$ at $y=-\sqrt{3}$

$\therefore$ point of triangle are

$(-3,0),(0,\sqrt{3}),(0,-\sqrt{3})$

$\Rightarrow D=\frac{1}{2}\left|\begin{array}{ccc}1& -3& 0\\ 1& 0& \sqrt{3}\\ 1& 0& -\sqrt{3}\end{array}\right|=(1\left(0\right)+3(-2\sqrt{3}))\frac{1}{2}$

$=3\sqrt{3}$