Let the equation x^{2}+y^{2}+px+(1-p)y+5=0 re | vedclass

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(B) The radius $r$ of the circle $x^{2}+y^{2}+px+(1-p)y+5=0$ is given by $r = \sqrt{(\frac{p}{2})^{2} + (\frac{1-p}{2})^{2} - 5} = \frac{\sqrt{p^{2} +

Vedclass · Accepted Answer

$61$

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(B) The radius $r$ of the circle $x^{2}+y^{2}+px+(1-p)y+5=0$ is given by $r = \sqrt{(\frac{p}{2})^{2} + (\frac{1-p}{2})^{2} - 5} = \frac{\sqrt{p^{2} + 1 - 2p + p^{2} - 20}}{2} = \frac{\sqrt{2p^{2} - 2p - 19}}{2}$.Since $r \in (0, 5]$,we have $0 Squaring gives $0 Solving $2p^{2} - 2p - 19 > 0$,the roots are $p = \frac{2 \pm \sqrt{4 + 152}}{4} = \frac{1 \pm \sqrt{39}}{2}$. So $p  \frac{1+\sqrt{39}}{2}$.Solving $2p^{2} - 2p - 19 \leq 100$,we have $2p^{2} - 2p - 119 \leq 0$. The roots are $p = \frac{2 \pm \sqrt{4 + 952}}{4} = \frac{1 \pm \sqrt{239}}{2}$. So $\frac{1-\sqrt{239}}{2} \leq p \leq \frac{1+\sqrt{239}}{2}$.Thus $p \in [\frac{1-\sqrt{239}}{2}, \frac{1-\sqrt{39}}{2}) \cup (\frac{1+\sqrt{39}}{2}, \frac{1+\sqrt{239}}{2}]$.Since $q = p^{2}$,$q$ ranges from $(\frac{1-\sqrt{39}}{2})^{2} \approx 7.7$ to $(\frac{1+\sqrt{239}}{2})^{2} \approx 68.7$ and from $(\frac{1-\sqrt{239}}{2})^{2} \approx 68.7$ to $(\frac{1+\sqrt{39}}{2})^{2} \approx 7.7$.Specifically,$q \in (7.7, 68.7]$. The integers $q$ are $8, 9, \dots, 68$. The number of elements is $68 - 8 + 1 = 61$.

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 $S = \{q : q = p^{2} \text{ and } q \text{ is an integer}\}$ is ..... .

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