Let the equation x^2+y^2+px+(1-p)y+5=0 represent circles of varying radius r(0,5].

Question

Let the equation $x^2+y^2+px+(1-p)y+5=0$ represent circles of varying radius $r\in(0,5]$. Then, the number of elements in the set $S=\{q:q=p^2	ext{ and }q	ext{ is an integer}\}$ is \_\_\_\_\_\_\_\_.

Answer: 61

Accepted Answer

$$r^2=\frac{p^2+(1-p)^2}4-5=\frac{2p^2-2p-19}4.$$

$0<r^2\le25\Rightarrow0<2p^2-2p-19\le100$.

Left: $p<\dfrac{1-\sqrt{39}}2$ or $p>\dfrac{1+\sqrt{39}}2$. Right: $\dfrac{1-\sqrt{239}}2\le p\le\dfrac{1+\sqrt{239}}2$.

So $p\in\left[\dfrac{1-\sqrt{239}}2,\dfrac{1-\sqrt{39}}2\right)\cup\left(\dfrac{1+\sqrt{39}}2,\dfrac{1+\sqrt{239}}2\right]$.

Computing $q=p^2$ at the four boundary points: $\dfrac{120-\sqrt{239}}2\approx52.27$, $\dfrac{20-\sqrt{39}}2\approx6.878$, $\dfrac{20+\sqrt{39}}2\approx13.12$, $\dfrac{120+\sqrt{239}}2\approx67.73$.

The two branches give $q\in(6.878,52.27]$ and $q\in(13.12,67.73]$, whose union is $(6.878,67.73]$. Integer values: $q=7,8,\dots,67$.

$67-7+1=61.$

Solutions · 1