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P and Q are two variable points on the axes of x and y respectively such that |OP|+|OQ|=α , then the locus of foot of perpendicular from (0, 0) on PQ is

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a

(x+y)(x2+y2)=a(x−y)

b

(x−y)(x2−y2)=axy

c

(x+y)(x2+y2)=axy

d

(x+y)(x2−y2)=axy

answer is C.

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Detailed Solution

Let PQ¯ be xα+yβ=1 equation of the circle passing through O, P, R is x2+y2−αx=0.⇒α=x2+y2x where Ris foot of perpendicularSimilarly β=x2+y2y ∴ |α|+|β|=aNow eliminate α, β

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