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प्रश्न
If the radius of the circle x2 + y2 + ax + (1 − a) y + 5 = 0 does not exceed 5, write the number of integral values a.
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उत्तर
According to the question, we have:
\[\sqrt{\left( \frac{- a}{2} \right)^2 + \left( \frac{a - 1}{2} \right)^2 - 5} \leq 5\]
\[ \Rightarrow \left( \frac{- a}{2} \right)^2 + \left( \frac{a - 1}{2} \right)^2 \leq 30\]
\[\Rightarrow a^2 + \left( a - 1 \right)^2 \leq 120\]
\[ \Rightarrow 2 a^2 - 2a - 119 \leq 0\]
Using quadratic formula:
\[ a = \frac{2 \pm \sqrt{2^2 - 4\left( 2 \right)\left( - 119 \right)}}{2\left( 2 \right)}\]
\[ \Rightarrow a = \frac{2 \pm \sqrt{956}}{4}\]
\[ \Rightarrow a = \frac{1 \pm 15 . 46}{2}\]
\[ \Rightarrow a = - 7 . 23, 8 . 23\]
\[ \Rightarrow - 7 . 23 \leq a \leq 8 . 23\]
\[ \Rightarrow a = - 7, - 6, - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5, 6, 7, 8 \left( If a \in \mathbb{Z} \right)\]
The number of integral values of a is 16.
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