Let and be two distinct points on a circle with center . Let O be the origin and OC be perpendicular to both CP and CQ. If the area of the triangle OCP is , then is equal to

Question

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 $\sqrt{35}$ , then $a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2}$ is equal to _________.

Smart Achievers · Accepted Answer

(66)

$Since area of ∆ O C P = \frac{1}{2} \times P C \times \sqrt{5} = \frac{\sqrt{35}}{2},$ $P C = \sqrt{7}$

$a_{1}^{2} + b_{1}^{2} + a_{2}^{2} + b_{2}^{2} = O P^{2} + O Q^{2} = 2 (5 + 7) = 24$

Solution

Ans.
(66)

$Since area of ∆ O C P = \frac{1}{2} \times P C \times \sqrt{5} = \frac{\sqrt{35}}{2},$ $P C = \sqrt{7}$

$a_{1}^{2} + b_{1}^{2} + a_{2}^{2} + b_{2}^{2} = O P^{2} + O Q^{2} = 2 (5 + 7) = 24$

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Solution

Q. Let P(a1,b1) and Q(a2,b2) be two distinct points on a circle with center C(2,3). Let O be the origin and OC be perpendicular to both CP and CQ. If the area of the triangle OCP is 35, then a12+a22+b12+b22 is equal to _________.

Ans. (66) Since area of ∆OCP=12×PC×5=352, PC=7 a12+b12+a22+b22=OP2+OQ2=2(5+7)=24

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Ans.
(66)

$Since area of ∆ O C P = \frac{1}{2} \times P C \times \sqrt{5} = \frac{\sqrt{35}}{2},$ $P C = \sqrt{7}$

$a_{1}^{2} + b_{1}^{2} + a_{2}^{2} + b_{2}^{2} = O P^{2} + O Q^{2} = 2 (5 + 7) = 24$