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This is the original phrasing of the question:

"Describe a circle to touch a given circle, and also to touch a given straight line at a given point."

  • A School Geometry H.S. Hall and F.H. Stevens p.184 q.9

See Figure 1. drawn in Geogebra

In the figure, suppose AB is the given line and the circle with center O and radius OP is the given circle.

This is part of a euclidean geometry textbook, so it would be preferred if the answer can be kept within the scope of euclidean geometry.

There are a few Theorems which I believe would help:

  1. If two circles touch one another, the centre, and the point of contact are in one straight line (Hence the centre must lie on a straight line from the given circle's centre). The difficulty is in ascertaining which radius is to be produced to find the centre of the required circle.

  2. For a circle to touch a straight line at a given point, the straight line must be perpendicular to a radius at that point. Hence the centre must lie on the perpendicular to the given straight line at the given point.

  3. Of course, radii of the same circle are equal. Hence PS = SR

I am quite stuck on this problem. Thank you very much for any help. They are greatly appreciated.

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  • $\begingroup$ If you can use hyperbolae, then: let $Q$ the center of the given circle, $R$ its radius, $t$ the tangent line, $P$ the tangency point, and $M$ the midpoint of $PQ$. Then the center of the required circle lies on the intersection between the line perpendicular to $t$ through $P$ and the hyperobla branch with focii $P$ and $Q$ and passing trhough $N \in PQ$, where $N$ lies between $M$ and $P$ and such that $MN \cong R/2$. $\endgroup$ Commented Nov 1, 2023 at 19:47
  • $\begingroup$ Yes. I believe that would work. But with these types of problems about geometric constructions, I'm mostly limited to straight lines and circles (drawn by a compass) in the textbook. $\endgroup$ Commented Nov 1, 2023 at 20:09

2 Answers 2

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Let the point on the line be $P$ and the center and the radius of the circle be $O$ and $r$, respectively.

  • Begin by drawing a perpendicular line to the given line at point $P$.

  • On this perpendicular line, locate and mark point $F$ at a distance of $r$ from point $P$.

  • Next, construct the perpendicular bisector of line segment $OF$.

  • The intersection point of the perpendicular bisector with line $PF$ will serve as the center for the circle in question.

It's worth noting that there are two possible positions for point $F$, resulting in the possibility of two different circles.

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  • $\begingroup$ I've drawn a rough sketch of your construction here: geogebra.org/geometry/bt7znbt6 Is this what you mean? $\endgroup$ Commented Nov 1, 2023 at 21:00
  • $\begingroup$ The required circle is supposed to be tangent to both the given circle and the given line at P. Hence the distance from the centre to both points of contact (for the tangents) should be equal. I can't seem to figure out how that factor s into the construction. $\endgroup$ Commented Nov 1, 2023 at 21:10
  • $\begingroup$ @procommania Yes. Try to draw circle with center at $D$ and radius $DP$. What do you see? $\endgroup$ Commented Nov 1, 2023 at 21:26
  • $\begingroup$ Brilliant. I didn't realize they are internally tangent. Thank you very much. $\endgroup$ Commented Nov 1, 2023 at 21:29
  • $\begingroup$ @procommania You are welcome! You can also try the second possible position of the point $F$. :) $\endgroup$ Commented Nov 1, 2023 at 22:50
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Let the given circle be centered at $C_1$ with a radius of $r_1$. And the let the line be given parametrically by

$ P(t) = P_0 + t \ d , t \in \mathbb{R}$

where we can assume without any loss of generality that $d$ is a unit vector.

where $P_0$ is the point where the circle to be constructed is tangent to the line.

The center $C_2$ lies on the perpendicular line to the line $P(t)$. Hence,

$ C_2 = P_0 + t \ n $

where $ n = [ - d_y , d_x ] $

Note that $n$ is also a unit vector.

To make sure we got the direction of $n$ right. We must have $ n \cdot (C_1 - P_0) $ positive. If it is not, then we'll just reverse the direction of $n$. Now we require that $t \gt 0$.

The equation that determines $t$ (which is also the radius) is

$ (t + r_1) = ( P_0 + t n - C_1) \cdot (P_0 + t n - C_1) $

Expanding the right hand gives us

$ t^2 + 2 t r_1 + r_1^2 = t^2 + 2 t \ n \cdot (P_0 - C_1) + (P_0 - C_1) \cdot (P_0 - C_1) $

Cancelling the term $t^2$ on both sides, and solving for $t$, we find that

$ t = \dfrac{ r_1^2 - (P_0 - C_1 ) \cdot (P_0 - C_1) }{ 2 (\ n \cdot (P_0 - C_1 ) - r_1)} $

Of course, $t = r_2$ the radius of the circle, and its center is

$ C_2 = P_0 + t \ n $

As a numerical example, suppose

$ C_1 = (3, 8)$

$ r_1 = 5 $

$P_0 = (-6, 4) $

$ d = \dfrac{1}{\sqrt{5}} (1, 2 ) $

Then applying the above, we get our circle as shown in the figure below. The given circle is drawn in black, and line in green, and the constructed circle is in red.

Constructed Circle

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