# Draw a circle of unknown radius using ruler-compass construction

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?

• This is not clear. What data is given? What do you want to construct?
– lulu
Jan 17, 2020 at 15:49
• @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. Jan 17, 2020 at 15:53
• 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 $B$. Jan 17, 2020 at 15:54
• @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. Jan 17, 2020 at 15:54
• @MoonLightSyzygy You are not supposed to do any computing. This is an exercise in construction.
– YNK
Jan 17, 2020 at 16:16

$$\bf\rm{Fig. 1}$$ 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!

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