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 = 4Rx2y2
(b)
(x2+y2)(x+y) = R2xy
(c)
(x2+y2)3 = 4R2x2y2
(d)
(x2+y2)2 = 4R2x2y2
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 = 4Rx2y2
(b)
(x2+y2)(x+y) = R2xy
(c)
(x2+y2)3 = 4R2x2y2
(d)
(x2+y2)2 = 4R2x2y2
Question ID - 57312 | SaraNextGen Top Answer
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 = 4Rx2y2
(b)
(x2+y2)(x+y) = R2xy
(c)
(x2+y2)3 = 4R2x2y2
(d)
(x2+y2)2 = 4R2x2y2
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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 |
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|
(a) |
(x2+y2)2 = 4Rx2y2 |
(b) |
(x2+y2)(x+y) = R2xy |
|
(c) |
(x2+y2)3 = 4R2x2y2 |
(d) |
(x2+y2)2 = 4R2x2y2 |
1 Answer
127 votes
Answer Key / Explanation : (c) -
Slope of AB = 
Equation of AB is hx+ky = h2 + k2
A
AB = 2R
⇒ (h2 + k2)3 = 4R2h2k2
⇒ (x2+y2)3 = 4R2x2y2
127 votes
Answer Link
127