18. The system $x^{2}-y^{2}=0,(x-a)^{2}+y^{2}=1$ has generally at most four solutions. Find the values of $a$ so that the system has two or three solutions.
(ans. $a= \pm 1$ for two solutions, $a= \pm \sqrt{2}$ for three solutions.
The solutions are given by $x=\frac{a \pm \sqrt{2-a^{2}}}{2}, y= \pm x$. There are two solutions if the quadratic equation involving $x$ has a single solution $\Rightarrow$ $2-a^{2}=0 \Rightarrow a= \pm \sqrt{2}$. If one value of $x$ is 0 , then there will at most be 3 solutions. Solving $x=0$ in $a$ yields $a= \pm 1$ and this gives exactly 3 solutions).