If tangents OQ and OR are drawn to variable circles having radius r and the centre lying on the rectangular hyperbola xy=1, then locus of circumcentre of triangle OQR is (O being the origin).
A
xy=4
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B
xy=14
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C
xy=1
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D
xy=−4
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Solution
The correct option is Bxy=14 Let S be any point on the rectangular hyperbola (t,1t)
Hence eqaution of the circle with radius r is (x−t)2+(y−1/t)2=r2
Equation of chord of contact from (0,0) will be −xt−yt+t2+1t2−r2=0
Family of circles equation passing through P,Q is (x−t)2+(y−1/t)2−r2+k(−xt−yt+t2+1t2−r2)=0
If it passes through (0,0) ⇒k=−1
Hence circumcircle equation for triangle OQR will be (x−t)2+(y−1/t)2−r2+xt+yt−t2−1t2+r2=0⇒x2+y2−xt−yt=0
whose centre is (t/2,1/2t)≡(h,k) ∴ required locus equation will be xy=14