A circle of radius $2$ units rolls on the outerside of the circle, $x^2 + y^2 + 4x = 0$ , touching it externally. Find the locus of the centre of this outer circle. Also find the equations of the common tangents of the two circles when the line joining the centres of the two circles is inclined at an angle of $60^\circ$ with $x$-axis

My attempt is as follows:-

First part of the question is simple, so I will directly write its answer.


Now in the second part I am confused they are talking about which two circles, are they talking about the circles given initially?

I had no choice so I took the circles given initially.

$$C_1\equiv(-g,-f)$$ $$C_2\equiv(-2,0)$$

Equation of line $C_1C_2$

$$y=\sqrt{3}(x+2)$$ $$y=x\sqrt{3}+2\sqrt{3}\tag{1}$$

Now as circles are touching externally, so $C_1C_2=r_1+r_2=4$

Let's find possible values of $C_2$ using parametric form of line

For $r=4, C_2\equiv(0,2\sqrt{3})$

For $r=-4, C_2\equiv(-4,-2\sqrt{3})$

Case $1$: $C_2\equiv(0,2\sqrt{3})$

Let's find the equation of internal tangent. Two circles are touching at point $\left(-1,\sqrt{3}\right)$

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

Now as $r_1=r_2$, so external tangent will be parallel to $C_1C_2$, hence of equation of external tangent will be $y-\sqrt{3}x+\lambda=0$

$$\dfrac{\left|\lambda+2\sqrt{3}\right|}{2}=2$$ $$\lambda=4-2\sqrt{3},-4-2\sqrt{3}$$

So external tangent will be $y-\sqrt{3}x+4-2\sqrt{3}=0,y-\sqrt{3}x-4-2\sqrt{3}=0$

Case $2$: $C_2\equiv(-4,-2\sqrt{3})$

Let's find the equation of internal tangent. Two circles will touch at $(-3,-\sqrt{3})$

$$-y\sqrt{3}-3x+2(x-3)=0$$ $$-x-\sqrt{3}y-6=0$$ $$x+\sqrt{3}y+6=0$$

External tangents will remain same.

Now actual answer does not contain the internal tangent of case $2$. Is that not valid?

  • $\begingroup$ There is only one case when angle between line that connect centers and axis $x$ is $60$ degrees, right? In your second case the angle is $240$ degrees. I do agree that it can be interpreted differently. $\endgroup$
    – Vasili
    Jan 17, 2020 at 17:16
  • $\begingroup$ The fixed circle and the locus that you determined are concentric, so it’s a safe bet that they’re talking about tangents to the circles given initially. As for your question at the end, the problem is worded ambiguously. Working backwards from the given solution, it would appear that what was actually meant the line through the two centers is a ray originating at the center of the fixed circle. $\endgroup$
    – amd
    Jan 17, 2020 at 20:25

1 Answer 1


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Some of the text you used to describe your answer is not entirely clear to me. Can you please compare the method posted bellow with yours and let me know whether you need more clarification? It is important that you read Vasya's comment before proceeding.

The line Joining the two centers makes an angle of $60^0$ with the $x$-axis. Since the two tangents are perpendicular to this line they make an angle of $150^0$ with the $x$-axis. therefore, the equations of tangents can be written as $$y=-\frac{1}{\sqrt{3}}x+d_1\tag{Tangent 1}$$ $$y=-\frac{1}{\sqrt{3}}x+d_2\tag{Tangent 2}$$

Now, to determine $d_1$ and $d_2$, we have to find the coordinates of the points $P_1$ and $P_2$ $$P_1 = \left(-2+2\cos60^0, 2\sin60^0\right)=\left(-1, \sqrt{3}\right)$$ $$P_2 = \left(-2-2\cos60^0, -2\sin60^0\right)=\left(-3, -\sqrt{3}\right)$$

When we substitute these values in the equations of the two tangents, we get $$d_1=\frac{2}{\sqrt{3}}$$ $$d_2=-2\sqrt{3}$$

Therefore the two equations are, $$\sqrt{3}y+x-2=0$$ $$\sqrt{3}y+x+6=0$$

Your answers are correct.


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