Distance between the centers of two circles

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In summary, the conversation discusses a translation project on "La leçon de Platon" by Dom Néroman, which deals with the relationship between music theory and mathematics in Plato's works. The problem at hand involves a mathematical diagram and the use of Pythagoras' theorem to find the locus of a point on a vertical line. The conversation also includes a discussion on the use of Tikz coding for diagrams.
  • #1
Kobzar
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Hello, everybody:

I am a philologist who is fond of mathematics, but who unfortunately has just an elementary high school knowledge of them. I am translating La leçon de Platon, by Dom Néroman (La Bégude de Mazenc, Arma Artis, 2002), which deals with music theory and mathematics in the works of Plato. The problem which brings me here is not about translation, but about mathematics. Please see attachment.

Thank you very much in advance for whatever answer, and best regards.
 

Attachments

  • Distance between centers.pdf
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  • #2
Reference the attached marked diagram.

Line segments PB and PA are tangent to each circle. As such, they form right angles $PAC$ and $PBC'$

Using Pythagoras ...

$PC^2 = R^2 + PA^2 \implies PA^2 = PC^2 - R^2$
$PC'^2 = R'^2 + PB^2 \implies PB^2 = PC'^2 - R'^2$

$PA^2 = PB^2 \implies PC^2 - R^2 = PC'^2 - R'^2$

from the last equation above ...

$R'^2 - R^2 = PC'^2 - PC^2$

using Pythagoras again ...

$R'^2 - R^2 = (PD^2+C'D^2) - (PD^2+CD^2)$

$R'^2 - R^2 = C'D^2 - CD^2$

factoring the right side ...

$R'^2 - R^2 = (C'D - CD)(C'D+CD)$

note $C'D = a - CD$ and $a = C'D + CD$

substituting ...

$R'^2 - R^2 = (a - 2CD)(a)$

$\dfrac{R'^2-R^2}{a} = a - 2CD$

$2CD = a - \dfrac{R'^2-R^2}{a}$

$2CD = \dfrac{a^2}{a} - \dfrac{R'^2-R^2}{a}$

$2CD = \dfrac{a^2 - (R'^2-R^2)}{a}$

$CD = \dfrac{a^2 - (R'^2-R^2)}{2a}$

circle_problem.jpg
 
  • #3
[TIKZ]\draw [very thick] circle (2) ;
\draw [very thick] (7,0) circle (3) ;
\draw [thick] (-2,0) -- (11,0) ;
\draw [thick] (3.14,-3) -- (3.14,4) ;
\coordinate [label=right:$A$] (A) at (1.95,-0.5) ;
\coordinate [label=above:$B$] (B) at (7.5,2.95) ;
\coordinate [label=left:$P$] (P) at (3.14,3.5) ;
\coordinate [label=above:$C$] (C) at (0,0) ;
\coordinate [label=above right:$C'$] (E) at (7,0) ;
\coordinate [label=above right:$D$] (D) at (3.14,0) ;
\draw [very thick] (P) -- node[ right ]{$d$} (1.65,-1.5) ;
\draw [very thick] (P) -- node[ above ]{$d$} (8.5,2.85) ;
\draw [very thick] (C) -- node[ below ]{$R$} (A) ;
\draw [very thick] (E) -- node[ right ]{$R'$} (B) ;
\draw (C) -- (P) -- (E) ;
\draw (1.5,0.2) node {$x$} ;
\draw (5,0.2) node {$a-x$} ;
\draw (3.4,1.7) node {$h$} ;[/TIKZ]

Write $d$ for the equal distances $PA$ and $PB$, $h$ for $PD$, and $x$ for $CD$, so that $DC = a-x$.

Since $PC$ is the hypotenuse of both of the right-angled triangles $PAC$ and $PDC$, it follows that $$R^2+d^2 = x^2+h^2.$$ Similarly, it follows from the triangles $PBC'$ and $PDC'$ that $R'^2 + d^2 = (a-x)^2 + h^2.$ Subtract the first of those equations from the second, to get $$R'^2-R^2 = (a-x)^2-x^2 = a^2 -2ax.$$ Therefore $R'^2 - R^2 = a^2-2ax$, so that $x = \dfrac{a^2 - (R'^2 - R^2)}{2a}.$

Notice that $x$ turns out to depend only on $R$, $R'$ and $a$, and is independent of $d$ and $h$. This shows that the locus of $P$ is indeed the vertical line through $D$.

Edit: having posted that, I see that skeeter got there first!
 
  • #4
Opalg said:
Edit: having posted that, I see that @skeeter got there first!

very nice diagram ... TIKZ?
 
  • #5
skeeter said:
very nice diagram ... TIKZ?
Yes, but don't copy my Tikz coding – it's full of clumsy kludges.
 
  • #6
Opalg said:
[TIKZ]\draw [very thick] circle (2) ;
\draw [very thick] (7,0) circle (3) ;
\draw [thick] (-2,0) -- (11,0) ;
\draw [thick] (3.14,-3) -- (3.14,4) ;
\coordinate [label=right:$A$] (A) at (1.95,-0.5) ;
\coordinate [label=above:$B$] (B) at (7.5,2.95) ;
\coordinate [label=left:$P$] (P) at (3.14,3.5) ;
\coordinate [label=above:$C$] (C) at (0,0) ;
\coordinate [label=above right:$C'$] (E) at (7,0) ;
\coordinate [label=above right:$D$] (D) at (3.14,0) ;
\draw [very thick] (P) -- node[ right ]{$d$} (1.65,-1.5) ;
\draw [very thick] (P) -- node[ above ]{$d$} (8.5,2.85) ;
\draw [very thick] (C) -- node[ below ]{$R$} (A) ;
\draw [very thick] (E) -- node[ right ]{$R'$} (B) ;
\draw (C) -- (P) -- (E) ;
\draw (1.5,0.2) node {$x$} ;
\draw (5,0.2) node {$a-x$} ;
\draw (3.4,1.7) node {$h$} ;[/TIKZ]

Write $d$ for the equal distances $PA$ and $PB$, $h$ for $PD$, and $x$ for $CD$, so that $DC = a-x$.

Since $PC$ is the hypotenuse of both of the right-angled triangles $PAC$ and $PDC$, it follows that $$R^2+d^2 = x^2+h^2.$$ Similarly, it follows from the triangles $PBC'$ and $PDC'$ that $R'^2 + d^2 = (a-x)^2 + h^2.$ Subtract the first of those equations from the second, to get $$R'^2-R^2 = (a-x)^2-x^2 = a^2 -2ax.$$ Therefore $R'^2 - R^2 = a^2-2ax$, so that $x = \dfrac{a^2 - (R'^2 - R^2)}{2a}.$

Notice that $x$ turns out to depend only on $R$, $R'$ and $a$, and is independent of $d$ and $h$. This shows that the locus of $P$ is indeed the vertical line through $D$.

Edit: having posted that, I see that skeeter got there first!
Thank you very much!
 
  • #7
Opalg said:
Yes, but don't copy my Tikz coding – it's full of clumsy kludges.
I usually look at the source of Opalg's pictures to help me get rid of my own clumsy kludges. ;)
 
  • #8
skeeter said:
Reference the attached marked diagram.

Line segments PB and PA are tangent to each circle. As such, they form right angles $PAC$ and $PBC'$

Using Pythagoras ...

$PC^2 = R^2 + PA^2 \implies PA^2 = PC^2 - R^2$
$PC'^2 = R'^2 + PB^2 \implies PB^2 = PC'^2 - R'^2$

$PA^2 = PB^2 \implies PC^2 - R^2 = PC'^2 - R'^2$

from the last equation above ...

$R'^2 - R^2 = PC'^2 - PC^2$

using Pythagoras again ...

$R'^2 - R^2 = (PD^2+C'D^2) - (PD^2+CD^2)$

$R'^2 - R^2 = C'D^2 - CD^2$

factoring the right side ...

$R'^2 - R^2 = (C'D - CD)(C'D+CD)$

note $C'D = a - CD$ and $a = C'D + CD$

substituting ...

$R'^2 - R^2 = (a - 2CD)(a)$

$\dfrac{R'^2-R^2}{a} = a - 2CD$

$2CD = a - \dfrac{R'^2-R^2}{a}$

$2CD = \dfrac{a^2}{a} - \dfrac{R'^2-R^2}{a}$

$2CD = \dfrac{a^2 - (R'^2-R^2)}{a}$

$CD = \dfrac{a^2 - (R'^2-R^2)}{2a}$

View attachment 11019
Thank you very much!
 
  • #9
Klaas van Aarsen said:
I usually look at the source of Opalg's pictures to help me get rid of my own clumsy kludges. ;)
You're too modest, Klaas – you're definitely the Tikz expert around here.
 
  • #10
Opalg said:
You're too modest, Klaas – you're definitely the Tikz expert around here.
Not sure about that, but at least I'm able to do something about it if for some reason TikZ does not work as expected.
 

What is the formula for finding the distance between the centers of two circles?

The formula for finding the distance between the centers of two circles is √[(x2-x1)^2 + (y2-y1)^2], where (x1, y1) and (x2, y2) are the coordinates of the centers of the circles.

Can the distance between the centers of two circles be negative?

No, the distance between the centers of two circles cannot be negative. It is always a positive value.

How is the distance between the centers of two circles related to the radii of the circles?

The distance between the centers of two circles is equal to the sum of their radii. This is known as the sum of the radii theorem.

What is the significance of the distance between the centers of two circles in geometry?

The distance between the centers of two circles is important in determining whether the circles intersect, are tangent, or are disjoint.

Can the distance between the centers of two circles be greater than the sum of their radii?

No, the distance between the centers of two circles cannot be greater than the sum of their radii. If it is, then the circles do not intersect or touch each other.

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