Let the equation x2 - y2 - px - 1 - py - 5 - 0 represent circles of varying radius r - 0- 5- Then the number of elements in the set S - q - q - p2 and q is an integer is -

R^2 = p^24+ (1 p)^24 5 ⇒ 0 lt; p^2 + (1 p)^2 20 ≤ 100 20 lt; p^2 + (1 p)^2 ≤ 120 p ∈( 1 √(239)2, 1 √(39)2 ) ∪( 1+√(39)2, 1+√(239)2 ) p^2 ∈ [7, 67] ...

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Standard XII

Mathematics

Obtaining Centre and Radius of a Circle from General Equation of a Circle

Let the equat...

Question

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

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Solution

$r^{2} = \frac{p^{2}}{4} + \frac{(1 - p)^{2}}{4} - 5 \Rightarrow 0 < p^{2} + (1 - p)^{2} - 20 \leq 100$
$20 < p^{2} + (1 - p)^{2} \leq 120$
$p \in (\frac{1 - \sqrt{239}}{2}, \frac{1 - \sqrt{39}}{2}) \cup (\frac{1 + \sqrt{39}}{2}, \frac{1 + \sqrt{239}}{2})$
$p^{2} \in [7, 67]$
Number of integral values = 61

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Let the equation x2+y2+px+(1–p)y+5=0 represent circles of varying radius r∈(0,5]. Then the number of elements in the set S={q:q=p2 and q is an integer} is .

r2=p24+(1−p)24−5⇒0<p2+(1−p)2−20≤100 20<p2+(1−p)2≤120 p∈(1−√2392,1−√392)∪(1+√392,1+√2392) p2∈[7,67] Number of integral values = 61

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