Euclidean Geometry - Construction of a Circle Tangent to Given Circle, Line, and Point on Line

Question

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.

Answer 1

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.

Answer 2

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.