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Let \mathrm{P\left(a_{1}, b_{1}\right)} and \mathrm{Q\left(a_{2}, b_{2}\right)} be two distinct points on a circle with center \mathrm{C(\sqrt{2}, \sqrt{3})}. Let 0 be the origin and \mathrm{OC} be perpendicular to both CP and CQ.
If the area of the triangle OCP is \frac{\sqrt{35}}{2},then a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2} is equal to __________.

Option: 1

24


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer



\mathrm{OC \; is \perp^{\mathrm{r}} to \: both \; CP\: \&\: CQ }

\Rightarrow \mathrm{PQ} is a Diameter
\mathrm{Area \; of \; \triangle \mathrm{OCP}=\frac{\sqrt{35}}{2}}
\frac{1}{2} \times \mathrm{CP} \times \mathrm{OC}=\frac{\sqrt{35}}{2}
\mathrm{CP} \times \sqrt{2+3}=\sqrt{35}
\mathrm{CP}=\sqrt{7} \Rightarrow$ radius $=\sqrt{7}

\mathrm{Now \: \mathrm{OP}^{2}=\mathrm{OC}^{2}+\mathrm{PC}^{2}}
\mathrm{a_{1}^{2}+b_{1}^{2}=2+3+7=12}

Similarly

\mathrm{OQ}^{2}=\mathrm{OC}^{2}+\mathrm{CQ}^{2}
\mathrm{a}_{1}^{2}+\mathrm{b}_{1}^{2}=2+3+7=12
\therefore \, \mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\mathrm{b}_{1}^{2}+\mathrm{b}_{2}^{2}=24
 

Posted by

avinash.dongre

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