The two points A and B in a plane are such that for all points P lies on circle satisfied PA PB=k, then k will not be equal to (Ans least positive integer)

Question

Accepted Answer

Correct answer: 1. 1JEE Main 2020 Let A= (a ,0 ) ,B= (-a ,0 ) ,P= ( , ) ∴ PA^2PB^2=k^2 ⇒ ( -a )^2+ ^2=k^2 [ ( +a )^2+ ^2 ] ∴ Locus is (x^2+y^2 ) (1-k^2 )-2a (1+k^2 )x+ (1-k^2 )a^2=0 This is a circle for k 1

$The two points A and B in a plane are such that for all points P lies on circle satisfied \frac{PA}{PB} = k, then k will not be equal to (Ans least positive integer)$

Solution

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