Question:

Step 1: Identify the given coordinates and relationships. Let $O$ be the origin $(0,0)$ and $C$ be the center of the circle $(\sqrt{2},\sqrt{3})$. $P(a_1,b_1)$ and $Q(a_2,b_2)$ are points on the circle. The vector $\vec{OC}$ is given by: $$ \vec{OC} = C - O = (\sqrt{2}, \sqrt{3}) $$ The magnitude of $OC$ is: $$ OC = |\vec{OC}| = \sqrt{(\sqrt{2})^2 + (\sqrt{3})^2} = \sqrt{2+3} = \sqrt{5} $$ Step 2: Use the perpendicularity condition $OC \perp CP$ and $OC \perp CQ$. Since $OC$ is perpendicular to $CP$, their dot product is zero. Let $R$ be the radius of the circle, so $CP = CQ = R$. The vector $\vec{CP}$ is $P-C = (a_1-\sqrt{2}, b_1-\sqrt{3})$. $$ \vec{OC} \cdot \vec{CP} = 0 $$ $$ \sqrt{2}(a_1-\sqrt{2}) + \sqrt{3}(b_1-\sqrt{3}) = 0 $$ $$ \sqrt{2}a_1 - 2 + \sqrt{3}b_1 - 3 = 0 $$ $$ \sqrt{2}a_1 + \sqrt{3}b_1 = 5 \quad \text{(Equation 1)} $$ Similarly, for point $Q(a_2,b_2)$, since $OC \perp CQ$: $$ \vec{OC} \cdot \vec{CQ} = 0 $$ $$ \sqrt{2}(a_2-\sqrt{2}) + \sqrt{3}(b_2-\sqrt{3}) = 0 $$ $$ \sqrt{2}a_2 - 2 + \sqrt{3}b_2 - 3 = 0 $$ $$ \sqrt{2}a_2 + \sqrt{3}b_2 = 5 \quad \text{(Equation 2)} $$ Step 3: Use the given area of $\triangle OCP$ to find the radius $R$. Since $OC \perp CP$, $\triangle OCP$ is a right-angled triangle with the right angle at $C$. The area of $\triangle OCP$ is given as $\dfrac{\sqrt{35}}{2}$. The area can also be expressed as $\dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times OC \times CP$. We know $OC = \sqrt{5}$ and $CP = R$. $$ \dfrac{1}{2} \times \sqrt{5} \times R = \dfrac{\sqrt{35}}{2} $$ $$ \sqrt{5} R = \sqrt{35} $$ $$ R = \frac{\sqrt{35}}{\sqrt{5}} = \sqrt{\frac{35}{5}} = \sqrt{7} $$ So, the square of the radius is $R^2 = 7$. Step 4: Relate $R^2$ to the coordinates of $P$. The point $P(a_1,b_1)$ lies on the circle with center $C(\sqrt{2},\sqrt{3})$ and radius $R=\sqrt{7}$. Therefore, $CP^2 = R^2 = 7$. $$ (a_1-\sqrt{2})^2 + (b_1-\sqrt{3})^2 = 7 $$ Expanding this equation: $$ a_1^2 - 2\sqrt{2}a_1 + (\sqrt{2})^2 + b_1^2 - 2\sqrt{3}b_1 + (\sqrt{3})^2 = 7 $$ $$ a_1^2 + b_1^2 - 2\sqrt{2}a_1 - 2\sqrt{3}b_1 + 2 + 3 = 7 $$ $$ a_1^2 + b_1^2 - 2(\sqrt{2}a_1 + \sqrt{3}b_1) + 5 = 7 $$ From Equation 1, we know $\sqrt{2}a_1 + \sqrt{3}b_1 = 5$. Substitute this value: $$ a_1^2 + b_1^2 - 2(5) + 5 = 7 $$ $$ a_1^2 + b_1^2 - 10 + 5 = 7 $$ $$ a_1^2 + b_1^2 - 5 = 7 $$ $$ a_1^2 + b_1^2 = 12 $$ Step 5: Relate $R^2$ to the coordinates of $Q$. Similarly, for point $Q(a_2,b_2)$, $CQ^2 = R^2 = 7$. $$ (a_2-\sqrt{2})^2 + (b_2-\sqrt{3})^2 = 7 $$ Expanding this equation: $$ a_2^2 - 2\sqrt{2}a_2 + 2 + b_2^2 - 2\sqrt{3}b_2 + 3 = 7 $$ $$ a_2^2 + b_2^2 - 2(\sqrt{2}a_2 + \sqrt{3}b_2) + 5 = 7 $$ From Equation 2, we know $\sqrt{2}a_2 + \sqrt{3}b_2 = 5$. Substitute this value: $$ a_2^2 + b_2^2 - 2(5) + 5 = 7 $$ $$ a_2^2 + b_2^2 - 10 + 5 = 7 $$ $$ a_2^2 + b_2^2 - 5 = 7 $$ $$ a_2^2 + b_2^2 = 12 $$ Step 6: Calculate the required sum. We need to find $a_1^2+a_2^2+b_1^2+b_2^2$. $$ a_1^2+a_2^2+b_1^2+b_2^2 = (a_1^2+b_1^2) + (a_2^2+b_2^2) $$ Substitute the values found in Step 4 and Step 5: $$ a_1^2+a_2^2+b_1^2+b_2^2 = 12 + 12 = 24 $$ The final answer is $\boxed{24}$.