Let $P (a_{1}, b_{1})$ and $Q (a_{2}, b_{2})$ be two distinct points on a circle with center $C (\sqrt{2}, \sqrt{3})$ . Let $O$ be the origin and $O C$ be perpendicular to both $C P$ and $C Q .$ If the area of the triangle $O C P$ is $\frac{\sqrt{35}}{2}$ , then $a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2}$ is equal to______.

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answer is 24.

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Detailed Solution

$\begin{array}{l} OC = \sqrt{5} \\ \frac{1}{2} \times CP \times OC = \frac{\sqrt{35}}{2} \Rightarrow CP = \sqrt{7} \\ {OP}^{2} = 5 + 7 = 12 \\ a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2} = {OP}^{2} + {OQ}^{2} = 12 + 12 = 24 \end{array}$

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