If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is :
(A)(x2+y2)2=4R2x2y2
(B)(x2+y2)3=4R2x2y2✓ Correct
(C)(x2+y2)2=4Rx2y2
(D)(x2+y2)(x+y)=R2xy
Answer: (B)
Solution
Let the foot of the perpendicular from O to the chord AB be (h,k). The chord then has equation hx+ky=h2+k2, meeting the axes at A(hh2+k2,0) and B(0,kh2+k2). Since AB is a diameter of length 2R, AB2=4R2 gives h2k2(h2+k2)3=4R2. Replacing (h,k) by (x,y), the locus is (x2+y2)3=4R2x2y2.
JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option