A) \[8+5\sqrt{3}\]
B) \[\frac{47+10\sqrt{6}}{2}\]
C) \[14+5\sqrt{3}\]
D) \[\frac{25+\sqrt{6}}{2}\]
Correct Answer: C
Solution :
\[{{(x-5/2)}^{2}}\frac{-25}{4}+{{(y-1/2)}^{2}}-1/4+5=0\] \[={{(x-5/2)}^{2}}+{{(y-1/2)}^{2}}=3/2\] on circle \[Q=5/2+\sqrt{3/2}\cos Q,\frac{1}{2}+\sqrt{3/2}\sin Q\] \[P{{Q}^{2}}={{\left( \frac{5}{2}+\sqrt{3/2}\cos Q \right)}^{2}}+{{\left( \frac{5}{2}+\sqrt{3/2}\sin Q \right)}^{2}}\]\[P{{Q}^{2}}=\frac{25}{2}+\frac{3}{2}+5\sqrt{3/2}(cos\,Q+tan\,Q)\] \[=14+5\sqrt{3/2}(cos\,Q+tan\,Q)\] \[\text{ma}{{\text{n}}^{\text{mr}}}=14+5\sqrt{3/2}(\sqrt{2})\]\[=14+5\sqrt{3}\]
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