Let equation x^{2}+y^{2}+p x+(1-p) y+5=0 represent circles of varying radius mathrm{r} ∈(0,5] . Th

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10-1.Circle and System of Circles

hard

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

$60$

$61$

$62$

$63$

(JEE MAIN-2021)

Solution

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

Since, $r \in(0,5]$

So, $0\,<\,2 \mathrm{p}^{2}-2 \mathrm{p}-19\, \leq 100$

$\Rightarrow \mathrm{p} \in\left[\frac{1-\sqrt{239}}{2}, \frac{1-\sqrt{39}}{2}\right) \cup\left(\frac{1+\sqrt{39}}{2}, \frac{1+\sqrt{239}}{2}\right]$ so, number of integral values of $\mathrm{p}^{2}$ is $61.$

Standard 11

Mathematics

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

Solution

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