A rod PQ of length 2a slides with its ends on the axes the locus of the circum-centre of Δ OPQ is
A
x2 + y2 = 2a2
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B
x2 + y2 = 4a2
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C
x2 + y2 = 3a2
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D
x2 + y2 = a2
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Solution
The correct option is Dx2 + y2 = a2 Let C (h, k) be a point in the locus
C(h, k) is mid point of PQ ⇒ P(2h, 0), Q(0, 2k)
PQ = 2a ⇒PQ2=4a2⇒4h2+4k2=4a2⇒h2+k2=a2
Locus of (h, k) is x2 + y2 = a2