Question

The triangle formed by the common tangents of the circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$ is:

• A. an isosceles triangle
• B. an equilateral triangle
• C. a scalene triangle
• D. a right angled triangle

Answer 1

Answer: (A)

an isosceles triangle

The given circle $S_1:x^2-y^2+2x=0$

$\Rightarrow S_1:(x+1{)}^2+y^2$

$C_1(-1,0)$ and Radius $R_1=1$

$S_2:x^2+y^2-6x=0$

$\Rightarrow S_2:(x-3{)}^2+y^2=3^2$

$C_2(3,0)$ and Radius $R_2=3$

Now, $△PQ{C}_1$ and $△PS{C}_2$ are similar

$\Rightarrow \frac{P{C}_1}{P{C}_2}=\frac{Q{C}_1}{Q{C}_2}=\frac{1}{3}$

$\Rightarrow \frac{P{C}_1}{P{C}_1+C_1{C}_2}=\frac{1}{3}$

$\Rightarrow \frac{P{C}_1}{P{C}_1+4}=\frac{1}{3}$

$\Rightarrow P{C}_1=2$

$\therefore$ The cordinates of $p(x,o)$ will be $(-3,0)$

$\therefore$ In $△PQ{C}_1,\sin \theta =\frac{Q{C}_1}{P{C}_1}=\frac{1}{2}$

$\Rightarrow \theta ={30}^o$

Equation of $PS=\frac{y-o}{x-(-3)}=\tan {30}^o=\frac{1}{\sqrt{3}}$

$\Rightarrow \sqrt{3}y=x+3$

Subsituting $x=0$

we get $A=$ The cordinate of $A(0,\sqrt{3})$

Similarly cordinate of $B(0,-\sqrt{3})$

In triangle $△PAB$

$PA=PB=3\sqrt{2}$

$AB=2\sqrt{3}$

$△PAB$ is an isosceles triangle