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Given two non-intersecting circles (i.e. their centers are separated by a distance larger than the sum of their radii), it is always possible to place a third circle on top of (or below) these circles such that the third circle is tangent to the first two circles. In fact there are infinitely many such third circles.

My question is: what is the curve described by the centers of such third circles?

If the parameters of the first two circles are $x_1, y_1, r_1$ and $x_2, y_2, r_2$ it seems to me that the center of the third circle $(x,y)$ should satisfy the following $$ \sqrt {(x-x_1)^2+(y-y_1)^2} - r_1 = \sqrt {(x-x_2)^2+(y-y_2)^2} - r_2$$ But the solution to this equation eludes me.

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  • $\begingroup$ Seems correct to me. The left side is the distance from $(x,y)$ to the first circle. The right side is the distance from $(x,y)$ to the second circle. Clearly, an $(x,y)$ that is the center of a circle tangent to both must satisfy this. Conversely, any $(x,y)$ which satisfies this equation admits a circle with a unique point of intersection with the first and second circles. $\endgroup$ – Dzoooks Jun 2 '18 at 20:49
  • $\begingroup$ @Dzoooks: Thanks for confirming my equation. Any thoughts on its solution, i.e. what is y = f(x)? $\endgroup$ – Jens Jun 2 '18 at 20:57
  • $\begingroup$ $y$ won't be a function of $x$, for example, when the two circles are $(x-2)^2 + y^2 = 1$ and $(x+2)^2 + y^2 = 1,$ since here your curve is the vertical line $x=0.$ $\endgroup$ – Dzoooks Jun 2 '18 at 21:00
  • $\begingroup$ @Dzoooks: Yes, a vertical straight line is possible when the two circles have equal radii, but what is the curve in general? A parabola? An ellipse? $\endgroup$ – Jens Jun 2 '18 at 21:03
  • $\begingroup$ It's a hyperbola. $\endgroup$ – Dzoooks Jun 2 '18 at 21:05
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HINT

Wlog we can assume

  • $x_1=y_1=y_2=0$
  • $r_1=1$
  • $x_2=a$
  • $r_2=r$

and consider

$$\sqrt {x^2+y^2} - 1 = \sqrt {(x-a)^2+(y)^2} - r$$

$$\sqrt {x^2+y^2} = \sqrt {(x-a)^2+y^2} +1- r$$

$${x^2+y^2} = (x-a)^2+y^2 +(1- r)^2+2(1-r) \sqrt {(x-a)^2+y^2}$$

$$x^2+y^2-x^2+2xa-a^2-y^2 -(1- r)^2= 2(1-r) \sqrt {(x-a)^2+y^2}$$

$$2xa-a^2 -(1- r)^2= 2(1-r) \sqrt {(x-a)^2+y^2}$$

$$[2xa-(a^2 +(1- r)^2)]^2= 4(1-r)^2 [(x-a)^2+y^2]$$

$$4x^2a^2-4xa(a^2 +(1- r)^2)+(a^2 +(1- r)^2)^2=4(1-r)^2x^2-8a(1-r)^2x+4a^2(1-r)^2+4(1-r)^2y^2$$

$$[4a^2-4(1-r)^2]x^2+[8a(1-r)^2-4a(a^2 +(1- r)^2]x-4(1-r)^2y^2=4a^2(1-r)^2-(a^2 +(1- r)^2)^2$$

Note that for $r=1$ we obtain

$$4a^2x^2-4a^3x=-a^4\implies 4x^2-4ax+a^2=0 \implies (2x-a)^2=0 \implies x=\frac a 2$$

For $r=2$ and $a =4$ we have

$$60x^2-240x-4y^2-225=0$$

enter image description here

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  • $\begingroup$ Thanks. But if I slog through the algebraics from where you stopped, how do I adjust the final equation for the cases where your assumption doesn't hold? $\endgroup$ – Jens Jun 2 '18 at 21:18
  • $\begingroup$ @Jens You can find the kind of curve and that result is true i general. If we need the equation for a general case we can always by translation a scaling to consider the simpler case. $\endgroup$ – gimusi Jun 2 '18 at 21:23
  • $\begingroup$ (+1) My understanding is that the curve is a hyperbola. I also see that if I rotate and scale the original parameters of my circles, I can arrive at your input assumptions and thus arrive at your final result. But how is your final result transformed to the hyperbola equation, i.e. $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$ $\endgroup$ – Jens Jun 2 '18 at 21:45
  • $\begingroup$ @Jens As you can see it is an hyperbola in the form $$\frac{(x-b)^2}{a^2}-\frac{y^2}{b^2} = 1$$ $\endgroup$ – gimusi Jun 2 '18 at 21:53
  • $\begingroup$ Got it. Thanks! $\endgroup$ – Jens Jun 2 '18 at 22:37

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