0
$\begingroup$

$C, C_1, C_2, C_3$ are three circles of radii $5, 3, 2, x$ units respectively. $C_1$ and $C_2$ touch each other externally and $C$ internally. $C_3$ touches $C_1$ and $C_2$ externally and $C$ internally.

Find the value of $x$.

I've figured out that the diagram looks like this. And I have a hunch that the centre of $C_3$ will be directly above the intersection point of $C_1$ and $C_2$. I've also tried constructing right-angled triangles and applying the pythagoras theorem.
enter image description here

$\endgroup$
  • $\begingroup$ Lookup Descartes' Theorem for the general case. $\endgroup$ – dxiv Mar 12 '17 at 17:35
  • $\begingroup$ The center $M$ of $C_3$ is at distance $5-x$ from the center of $C$ hence $M=(5+(5-x)\cos t,(5-x)\sin t)$ for some $t$. The distance from $M$ to the center $(3,0)$ of $C_1$ is $3+x$ and the distance from $M$ to the center $(8,0)$ of $C_2$ is $2+x$, hence $$(5+(5-x)\cos t-3)^2+((5-x)\sin t)^2=(3+x)^2$$ and $$(5+(5-x)\cos t-8)^2+((5-x)\sin t)^2=(2+x)^2$$ Can you finish this? $\endgroup$ – Did Mar 12 '17 at 17:46
2
$\begingroup$

(see picture below)

Let us denote, on homogeneity grounds, $C_0=C$ and by $C_{ij}$ the contact point of circle $C_i$ with circle $C_j$.

A common feature is that points $C_i,C_j$ and $C_{ij}$ are aligned. More precisely, we are interested by the following alignements:

  • $C_1,C_{13},C_3$ are aligned in this order, thus $C_1C_3=3+x$ (external contact).

  • $C_2,C_{23},C_3$ are aligned in this order, thus $C_2C_3=2+x$ (external contact).

  • $C_0,C_3,C_{03}$ are aligned in this order, thus $C_2C_3=5-x$ (internal contact).

Let $(x_3,y_3)$ be the coordinates of $C_3$.

We can write the following system of 3 equations with 3 unknowns $x,x_3,y_3$:

$$\begin{cases}\text{square of length} \ C_1C_3: & \ \ (x_3-3)^2+y_3^2&=&(3+x)^2 & \ \ (a)\\\text{square of length} \ C_2C_3: & \ \ (x_3-8)^2+y_3^2&=&(2+x)^2&\ \ (b)\\ \text{square of length} \ C_0C_3: & \ \ (x_3-5)^2+y_3^2&=&(5-x)^2&\ \ (c) \\ \end{cases}$$

Expanding and subtracting (a) from (b) and (b) from (c) we obtain a linear system with $x$ and $x_3$ as unknowns, giving

$$x=30/19 \approx 1.579$$

together with $x_3=120/19$ and, after plugging these results into (a): $y_3=60/19$.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.