Question

Equation of a common tangent to the circles $x^2+y^2-6x=0$ and $x^2+y^2+2x=0$ is

• A. $x=1$
• B. $x=0$
• C. $x+\sqrt{3}y+3=0$
• D. $x-\sqrt{3}y+3=0$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $x=0$
C $x+\sqrt{3}y+3=0$
D $x-\sqrt{3}y+3=0$

Equations of the given circles can be written as $(x-3{)}^2+y^2=3^2$ ...(1)
and $(x+1{)}^2+y^2=1^2$ ...(2)

Equation of any tangent to circle (2) is

$(x+1)\cos \theta +y\sin \theta =1$ ....(3)

This will be a tangent to circle (1) also if

$\frac{(3+1)\cos \theta -1}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}=\pm 3\Rightarrow 4\cos \theta -1=\pm 3$

That is, $\cos \theta =1$ or $\cos \theta =-\frac{1}{2}$

When $\cos \theta =1$, we have $\sin \theta =0$ and the equation of the common tangent becomes

$x+1=1\Rightarrow x=0$

When $\cos \theta =-\frac{1}{2},$ we have $\sin \theta =\pm \frac{\sqrt{3}}{2},$ and the equations of the common tangents are

$-\frac{1}{2}(x+1)\pm \frac{\sqrt{3}}{2}y=1\Rightarrow x-\sqrt{3}y+3=0$ and $x+\sqrt{3}y+3=0$