Circle Question 7
Question 7 - 30 January - Shift 2
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 $OC$ 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
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Answer: 24
Solution:
Formula: Area of Right angle triangle
$\frac{1}{2} \times PC \times \sqrt{5}=\frac{\sqrt{35}}{2} $
$ PC=\sqrt{7}$
$ \begin{aligned} & a_1^{2}+b_1^{2}+a_2^{2}+b_2^{2}=OP^{2}+OQ^{2} \\ & =2(5+7)=24 \end{aligned} $
