Mathematics - Circle Question with Solution | TestHub

$r=\sqrt{\frac{p^2}{4}+\frac{{\left(1-p\right)}^2}{4}-5}$

$=\frac{\sqrt{2p^2-2p-19}}{2}$

Since $r\in (0,5]$

So, $0<\frac{\sqrt{2p^2-2p-19}}{2}\le 5$

$0<2p^2-2p-19\le 100$

$\Rightarrow 2p^2-2p-19>0 ...\left(i\right)$

For, $2p^2-2p-19>0$

$p=\frac{2\pm \sqrt{4-4\times 2\times \left(-19\right)}}{4}$

$=\frac{2\pm \sqrt{156}}{4}$

$\Rightarrow \left(p-\frac{1+\sqrt{39}}{2}\right)\left(p-\frac{1-\sqrt{39}}{2}\right)>0....\left(ii\right)$

$p^2=10\pm \frac{\sqrt{39}}{2}$

Also, $2p^2-2p-119\le 0 ...\left(iii\right)$

For, $2p^2-2p-119=0$

$p=\frac{2\pm \sqrt{4-4\times 2\times \left(-119\right)}}{4}$

$=\frac{1\pm \sqrt{239}}{2}$

$\Rightarrow \left(p-\frac{1+\sqrt{239}}{2}\right)\left(p-\frac{1-\sqrt{239}}{2}\right)\le 0...\left(iv\right)$

$p^2=60\pm \frac{\sqrt{239}}{2}$

So, from $\left(ii\right) \& \left(iv\right)$

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

$\Rightarrow p^2\in \left[\left(10-\frac{\sqrt{39}}{2},60-\frac{\sqrt{239}}{2}\right)\cup \left(10+\frac{\sqrt{39}}{2},60+\frac{\sqrt{239}}{2}\right)\right]$

So, number of integral values of $p^2$ is $61$.