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Q.

A circle of constant radius a passes through the origin 0 and cuts the axes of coordinates at points P and Q. Then the equation of the locus of the foot of perpendicular from O to PQ is

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a

x2+y21x2+1y2=4a2

b

x2+y221x2+1y2=a2

c

x2+y221x2+1y2=4a2

d

x2+y21x2+1y2=a2

answer is C.

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Detailed Solution

The equation of line PQ isy−k=−hk(x−h)or hx+ky=h2+k2∴ Q≡h2+k2h,0 and P≡0,h2+k2kAlso, 2a=x12+y12or x12+y12=4a2Eliminating x1 and y1we havex2+y221x2+1y2=4a2
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A circle of constant radius a passes through the origin 0 and cuts the axes of coordinates at points P and Q. Then the equation of the locus of the foot of perpendicular from O to PQ is