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QJEEM23Ma0293: JEE Main 2023 Mathematics - Class 11 JEEM23Ma0293 Solution: Let and be two distinct points on a circle with center . Let be …

Let P(a1,b1)P\left(a_1, b_1\right) and Q(a2,b2)Q\left(a_2, b_2\right) be two distinct points on a circle with center C(2,3)C(\sqrt{2}, \sqrt{3}) . Let O\mathrm{O} be the origin and OC\mathrm{OC} be perpendicular to both CP\mathrm{CP} and CQ\mathrm{CQ} . If the area of the triangle OCP\mathrm{OCP} is 352\frac{\sqrt{35}}{2} , then a12+a22+b12+b22a_1^2+a_2^2+b_1^2+b_2^2 is equal to :

Step-by-Step Verified Solution

OC ⊥ CPOC \,\bot \,CP and OC ⊥ CQOC \,\bot \, CQ

⇒PCQ\Rightarrow PCQ is a straight line

JEE Main 2023 (Online) 30th January Evening Shift Mathematics - Circle Question 43 English Explanation

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

Let CP=CQ=ICP = CQ = I

[OCP]=12×OC×I=352[OCP] = {1 \over 2} \times OC \times I = {{\sqrt {35} } \over 2}

I=7I = \sqrt 7

OP=OQ=(OC)2+I2=5+7=12OP = OQ = \sqrt {{{(OC)}^2} + {I^2}} = \sqrt {5 + 7} = \sqrt {12}

a12+a22+b12+b22=(a12+b22)+(a22+b22)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)

OP2+OQ2=12+12=24O{P^2} + O{Q^2} = 12 + 12 = 24

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