Construction of a circle that touches given elements

Question

Two congruent circles that touch at point H are given. Let the line p be their common tangent that doesn't pass through the point H. Construct a circle that touches both given circles and the line p.

The most intuitive thing was to construct an equilateral triangle NOL (such that L is on p, and N and O are on the given circles) and find its circumscribed circle. However, the obtained circle intersects given circles at N and O, instead of just touching them. What am I doing wrong?

Answer 1

Why not using Descartes theorem; If circles with radii $r_1$. $r_2$ and $r_3$ are mutually tangent, then the radius of the circle touching all can be found by this relation:

$(\frac 1{r_1}+\frac1 {r_2}+\frac 1 {r_3}+\frac 1 R)^2=2 (\frac 1{r_1^2}+\frac1 {r_2^2}+\frac 1 {r_3^2}+\frac 1 {R^2})$

where R is the radius of circle to be found. We have:

$r_1=r_2=r$

$r_3=\infty$

which represent the line p. Putting these vales in relation we get:

$LP=R=\frac r4$

Now it is easy to construct the circle: Draw perpendicular HP from H to p. Draw a circle center on P with radius $\frac r 4$, it intersect HP at point C which is the center of the circle with radius R.Draw circle radius R center at C.

Answer 2

Let the radius of the "big" circles be $r$.

It is fairly easy to show that if you pick $P$ such that $|LP|=\frac{1}{4}r$, and draw the small circle with the centre $P$ and radius $\frac{1}{4}r$, it will touch both the big circles and the line.

That is because this ensures that the right-angled triangle $\triangle PHG$ has the ratio of sides making the right angle of $|PH|:|GH|=3:4$. Therefore, it is one of those $3:4:5$ right-angled triangles, and $|GP|=\frac{5}{4}r=r+\frac{1}{4}r$. In other words, the distance of the centres $G$ and $P$ is exactly equal to the sum of the radii of the circles, and so those circles touch. A symmetrical argument applies to the other big circle.

Now the construction should be immediately obvious.