Let the equation x ^ 2 + y ^ 2 + px + ( 1 - p ) y + 5 = 0 represent...

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

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

Since $r \in ( 0, 5 ]$

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

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

$⇒ 2{p}^{2} - 2p - 19 > 0 . . . \left(i\right)$

For, $2{p}^{2} - 2p - 19 > 0$

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

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

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

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

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

For, $2{p}^{2} - 2p - 119 = 0$

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

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

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

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

So, from \left(ii\right) \text{ and } \left(iv\right)$</p><p>$⇒ p \in \left[\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)\right]$</p><p>$⇒ {p}^{2} \in \left[\left(10 - \dfrac{\sqrt{39}}{2}, 60 - \dfrac{\sqrt{239}}{2}\right) \cup \left(10 + \dfrac{\sqrt{39}}{2}, 60 + \dfrac{\sqrt{239}}{2}\right)\right]$</p><p>So, number of integral values of${p}^{2} is $61$.