17. The diagram shows eight circles of two different sizes. The circles are arranged in concentric pairs so that the centres form a square. Each larger circle touches one other larger circle and two smaller circles. The larger circles have radius 1 .
What is the radius of each smaller circle?
A $\frac{1}{3}$
B $\frac{2}{5}$
C $\sqrt{2}-1$
D $\frac{1}{2}$
E $\frac{1}{2} \sqrt{2}$
Solution
C
Let the centres of the circles be $P, Q, R$ and $S$, as shown.
Let the radius of the smaller circles be $r$.
Then the length of $P Q$ is the sum of the radius of a larger circle and the radius of a smaller circle. That is, $P Q=1+r$. Similarly $Q R=1+r$. Also, $P R$ is the sum of the radii of two of the large circles. That is, $P R=1+1=2$.

Because $P Q R S$ is a square, $\angle P Q R$ is a right angle.
Therefore, applying Pythagoras' theorem to the triangle $P Q R$, we have
$$
(1+r)^{2}+(1+r)^{2}=2^{2} \text {. }
$$

It follows that
$$
2(1+r)^{2}=4
$$
and hence
$$
(1+r)^{2}=2
$$

Therefore
$$
1+r= \pm \sqrt{2}
$$
and hence
$$
r= \pm \sqrt{2}-1
$$

Because a radius must be positive, we deduce that the radius of each smaller circle is
$$
\sqrt{2}-1 \text {. }
$$