question_answer

A circle is inscribed into a rhombus ABCD with one angle \[60{}^\circ \]. The distance from the centre of the circle to the nearest vertex is equal to 1. If P is any point of the circle, then \[{{\left| PA \right|}^{2}}+{{\left| PB \right|}^{2}}+{{\left| PC \right|}^{2}}+{{\left| PD \right|}^{2}}\] is equal to

A) 9

B) 10     

C) 11

D) 12

Correct Answer: C

Solution :

\[r=\sqrt{3}\sin 30{}^\circ =\frac{\sqrt{3}}{2}\]

\[{{\left| PA \right|}^{2}}+{{\left| PB \right|}^{2}}+{{\left| PC \right|}^{2}}+{{\left| PD \right|}^{2}}\]

\[\Rightarrow \,{{\left( x-\sqrt{3} \right)}^{2}}+{{y}^{2}}+{{x}^{2}}+{{\left( y-1 \right)}^{2}}\]\[+{{\left( x+\sqrt{3} \right)}^{2}}+{{y}^{2}}+{{x}^{2}}+{{\left( y+1 \right)}^{2}}\]

\[=4\left( {{x}^{2}}+{{y}^{2}}+2 \right)\]

\[=4({{r}^{2}}+2)\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because \,\,\,{{x}^{2}}+{{y}^{2}}={{r}^{2}} \right]\]

\[=4\left( \frac{3}{4}+2 \right)=11\]