Given data

Center of the circle is $C(\sqrt{2},\sqrt{3}).$ Origin is $O(0,0)$.

Also, $OC$ is perpendicular to both $CP$ and $CQ$.

Interpret the perpendicular condition

Since $P$ and $Q$ lie on the circle centered at $C$, the vectors $\overrightarrow{CP}$ and $\overrightarrow{CQ}$ are radii.

Given $OC \perp CP \quad \text{and} \quad OC \perp CQ,$ both radii are perpendicular to the same line $OC$.

Through point $C$, there are exactly two opposite radius directions perpendicular to $OC$. Hence $P$ and $Q$ are endpoints of a diameter perpendicular to $OC$.

So if the radius is $r$, then $CP=CQ=r.$

Find the length of $OC$

$OC=\sqrt{(\sqrt{2})^2+(\sqrt{3})^2}=\sqrt{2+3}=\sqrt{5}.$

Use area of triangle $OCP$

Since $OC \perp CP$, triangle $OCP$ is right-angled at $C$.

Therefore, \text{Area}(\triangle OCP)=\frac{1}{2}\cdot OC\cdot CP=rac{1}{2}\cdot \sqrt{5}\cdot r.

Given area is $\frac{\sqrt{35}}{2}.$

Hence, $\frac{1}{2}\sqrt{5}\,r=\frac{\sqrt{35}}{2}$ $\sqrt{5}\,r=\sqrt{35}$ $r=\frac{\sqrt{35}}{\sqrt{5}}=\sqrt{7}.$

Use midpoint relation for opposite endpoints of a diameter

Since $P$ and $Q$ are diametrically opposite with center $C$, their midpoint is $C$.

Thus, $\left(\frac{a_1+a_2}{2},\frac{b_1+b_2}{2}\right)=(\sqrt{2},\sqrt{3}).$

Also, $CP=CQ=r=\sqrt{7}.$

Required quantity

We need $a_1^2+a_2^2+b_1^2+b_2^2=OP^2+OQ^2.$

Let $\vec{OP}=\vec{OC}+\vec{CP}, \qquad \vec{OQ}=\vec{OC}+\vec{CQ}.$

Since $P$ and $Q$ are opposite endpoints of a diameter, $\vec{CQ}=-\vec{CP}.$

Therefore, $OP^2=|\vec{OC}+\vec{CP}|^2=OC^2+CP^2+2\vec{OC}\cdot\vec{CP},$ $OQ^2=|\vec{OC}-\vec{CP}|^2=OC^2+CP^2-2\vec{OC}\cdot\vec{CP}.$

Adding, $OP^2+OQ^2=2OC^2+2CP^2.$

Now, $OC^2=5, \qquad CP^2=r^2=7.$

So, $OP^2+OQ^2=2(5)+2(7)=10+14=24.$

Final answer

$a_1^2+a_2^2+b_1^2+b_2^2=24.$