Let P(a_1, b_1) and Q(a_2, b_2) be two distin | vedclass

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(B) The center of the circle is $C(\sqrt{2}, \sqrt{3})$. The distance $OC$ is given by $OC = \sqrt{(\sqrt{2})^2 + (\sqrt{3})^2} = \sqrt{2 + 3} = \sqrt

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) The center of the circle is $C(\sqrt{2}, \sqrt{3})$. The distance $OC$ is given by $OC = \sqrt{(\sqrt{2})^2 + (\sqrt{3})^2} = \sqrt{2 + 3} = \sqrt{5}$.Since $OC \perp CP$,the triangle $OCP$ is a right-angled triangle at $C$.The area of $	riangle OCP = \frac{1}{2} 	imes OC 	imes CP = \frac{\sqrt{35}}{2}$.Substituting $OC = \sqrt{5}$,we get $\frac{1}{2} 	imes \sqrt{5} 	imes CP = \frac{\sqrt{35}}{2}$,which implies $CP = \sqrt{7}$.Since $P$ and $Q$ lie on the circle with center $C$ and radius $R = CP = \sqrt{7}$,we have $(a_1 - \sqrt{2})^2 + (b_1 - \sqrt{3})^2 = 7$ and $(a_2 - \sqrt{2})^2 + (b_2 - \sqrt{3})^2 = 7$.Expanding these,$a_1^2 + b_1^2 - 2\sqrt{2}a_1 - 2\sqrt{3}b_1 + 5 = 7 \implies a_1^2 + b_1^2 = 2 + 2\sqrt{2}a_1 + 2\sqrt{3}b_1$.Similarly,$a_2^2 + b_2^2 = 2 + 2\sqrt{2}a_2 + 2\sqrt{3}b_2$.Since $OC \perp CP$ and $OC \perp CQ$,the vectors $\vec{CP}$ and $\vec{CQ}$ are perpendicular to $\vec{OC} = (\sqrt{2}, \sqrt{3})$.Thus,$\sqrt{2}(a_1 - \sqrt{2}) + \sqrt{3}(b_1 - \sqrt{3}) = 0 \implies \sqrt{2}a_1 + \sqrt{3}b_1 = 5$.Similarly,$\sqrt{2}a_2 + \sqrt{3}b_2 = 5$.Then $a_1^2 + b_1^2 = 2 + 2(5) = 12$ and $a_2^2 + b_2^2 = 12$.Therefore,$a_1^2 + a_2^2 + b_1^2 + b_2^2 = 12 + 12 = 24$.

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