JEE MAIN - Mathematics (2023 - 30th January Evening Shift - No. 18)
Let $P\left(a_1, b_1\right)$ and $Q\left(a_2, b_2\right)$ be two distinct points on a circle with center $C(\sqrt{2}, \sqrt{3})$. Let $\mathrm{O}$ be the origin and $\mathrm{OC}$ be perpendicular to both $\mathrm{CP}$ and $\mathrm{CQ}$. If the area of the triangle $\mathrm{OCP}$ is $\frac{\sqrt{35}}{2}$, then $a_1^2+a_2^2+b_1^2+b_2^2$ is equal to :
Answer
24
Explanation
$$OC \,\bot \,CP$$ and $$OC \,\bot \, CQ$$
$$\Rightarrow PCQ$$ is a straight line
_30th_January_Evening_Shift_en_18_1.png)
$$OC = \sqrt {{{(\sqrt 2 )}^2} + {{(\sqrt 3 )}^2}} = \sqrt 5 $$
Let $$CP = CQ = I$$
$$[OCP] = {1 \over 2} \times OC \times I = {{\sqrt {35} } \over 2}$$
$$I = \sqrt 7 $$
$$OP = OQ = \sqrt {{{(OC)}^2} + {I^2}} = \sqrt {5 + 7} = \sqrt {12} $$
$$a_1^2 + a_2^2 + b_1^2 + b_2^2 = \left( {a_1^2 + b_2^2} \right) + \left( {a_2^2 + b_2^2} \right)$$
$$O{P^2} + O{Q^2} = 12 + 12 = 24$$
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