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Imagine a circle A of unknown radius, its tangent line passes through another circle B center. The distance between A's point of tangency and B's center is known (but not measured). B's radius is also known (but not measured). Circle B and the tangent line are already drawn on the paper.

Using only a compass and straight edge, how do I draw circle A such that it just touches circle B?

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  • $\begingroup$ This is not clear. What data is given? What do you want to construct? $\endgroup$
    – lulu
    Jan 17, 2020 at 15:49
  • $\begingroup$ @lulu I want to create/draw circle A so that it fits the descriptions above. The aformentioned distance and circle B's radius is defined on the paper. $\endgroup$ Jan 17, 2020 at 15:53
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    $\begingroup$ If $x$ is the radius of $A$, $d$ is the distance between the point of tangency of $A$ and $B$'s center and $r$ the radius of $B$, then by Pythagoras $x^2+d^2=(x+r)^2$. Therefore, $x=\frac{a^2-r^2}{2r}=\frac{(a+r)(a-r)}{2r}$. So, you only need to compute the fourth proportional of $a+r, a-r$ and $2r$. You can do this by drawing Thales' theorem picture for these three lengths. Once you have $x$ extend a radius of $B$ beyond the circle $x$ units and there is the center of $A$. There is one solution for each radius of $B$. $\endgroup$ Jan 17, 2020 at 15:54
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    $\begingroup$ @lulu - I think you want to draw a circle A tangential to B which has the property that one of A's tangent lines passes through B's center with the given distance. $\endgroup$
    – Henry
    Jan 17, 2020 at 15:54
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    $\begingroup$ @MoonLightSyzygy You are not supposed to do any computing. This is an exercise in construction. $\endgroup$
    – YNK
    Jan 17, 2020 at 16:16

1 Answer 1

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enter image description here

$\bf\rm{Fig. 1}$ shows what is given. $O_B$ is the center of the circle-B and $T$ is the point of tangency. The line $O_BT$ meets the circumference of the circle-B at $E$. It is assumed that the radius of the circle-B = $b$ and $O_bT=d$.

As MoonLightSyzygy pointed out in one of his comments, there are two different solutions to your problem, one is when $d\gt b$ and the other is when $d\lt b$. The difference is elucidated by the fact that the circles A and B of the former solution are touching externally and those of the latter touche each other internally. However, the 6-step recipe of the construction for both cases are the same. I also guarantee that the solutions given below are pure constructions and include no computing whatsoever.

$\bf\rm{Fig. 2}$ illustrates the construction of externally contacting circles in its entirety.

$\bf\rm{Step\space 1:}$ Draw the perpendicular to $TO_B$ at $T$. The center of the circle-A lies on this line.

$\bf\rm{Step\space 2:}$ Extend $TO_B$ to meet the circle-B again at $F$. Note that $TF=d+b$.

$\bf\rm{Step\space 3:}$ Draw an arc with radius $TE$ and center $T$ to intersect the perpendicular constructed
$\space\space\space\space\space\space\space\space\space\space\space\space\space$ in $\rm{Step\space 1}$ at $G$. Note that $TG=d-b$.

$\bf\rm{Step\space 4:}$ Draw a straight line joining $F$ and $G$ to cut the circlr-B at $H$

$\bf\rm{Step\space 5:}$ Join $O_B$ to $H$ and then extend this line to meet the perpendicular constructed
$\space\space\space\space\space\space\space\space\space\space\space\space\space$ in $\rm{Step\space 1}$ at $O_A$.

$\bf\rm{Step\space 6:}$ Draw the circle with radius $O_AH$ and center at $O_A$. This is the circle-A.

Since the case $d\lt b$ is construction-wise the same (see $\bf\rm{Fig. 3}$), it is not described here. I would like to let you figure out what happens when $d=b$. Now, the story does not end here, because someone (not me) has to rake his or her brain to prove the construction I described above. Happy hunting!

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  • $\begingroup$ @Gregory Leo Have you been able to prove or find a proof of the construction? $\endgroup$
    – YNK
    Jan 23, 2020 at 15:20

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