• question_answer The greatest integral value of a such that $\sqrt{9-{{a}^{2}}+2x-{{x}^{2}}}\ge \sqrt{16-{{x}^{2}}}$ for at least one positive value of x is A) $3$                       B)        $4$                       C) $6$                       D)        $7$

[d] We have $y=\sqrt{9-{{a}^{2}}+2ax-{{x}^{2}}}$ $\Rightarrow \,\,\,\,\,\,{{y}^{2}}+{{(x-a)}^{2}}=9$ And    $y=\sqrt{16-{{x}^{2}}}$           $\Rightarrow \,\,\,\,\,\,\,{{y}^{2}}+{{x}^{2}}=16$ The given equations represent semicircles. If  $a>7,$ then no solution exists. Thus, $a\le 7$.\ Hence, maximum value of a is 7.
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