MHB Distance between the centers of two circles

[Kobzar]

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.

 

[skeeter]

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 ...

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

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

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

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

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

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

 

[Opalg]

[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!

 

[skeeter]

Edit: having posted that, I see that @skeeter got there first!

very nice diagram ... TIKZ?

 

[Opalg]

very nice diagram ... TIKZ?

Yes, but don't copy my Tikz coding – it's full of clumsy kludges.

 

[Kobzar]

[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!

 

[I like Serena]

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. ;)

 

[Kobzar]

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!

 

[Opalg]

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.

 

[I like Serena]

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.