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A rod of fixed length ' l ' is sliding along the co-ordinate axes . Find the locus of the midpoint of the rod |
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Answer» Two rods of LENGTHS a and b slide along coordinate axes such that their ENDS are concyclic. Locus of the center of the circle isA4(x 2 +y 2 )=a 2 +b 2 B4(x 2 +y 2 )=a 2 −b 2 C4(x 2 −y 2 )=a 2 −b 2 DFor the rod of length bthe extremes point (o, q) and(o, q + b) and for the rod of length a the EXTREME points(p,o) and (p, p + a). The center ofcircle will be INTERSECTION betweenthe ⊥ eq bisectors of rodswith length a and b.(x,y)=(p+ 2a ,q+ 2b )Sox=p+ 2a y=q+ 2b x−p= 2a y−q= 2b __(1)Now we need radiusBut distance between (x, y) and(p,o) and also distance between(x,y) and (o,q) so(x−p) 2 +y 2 =x 2 +(y−q) 2 __(2)Now( 2a ) 2 +y 2 =x 2 +( 2b ) 2 4a 2 −b 2 =x 2 −y 2 a 2 −b 2 =4(x 2 −y 2 ) |
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