Two rods of lengths a and b slide along the axes in such a way that their ends are concyclic. The locus of the centre of the circle passing through these points is

Two rods of lengths a and b slide along the axes in such a way that th

1.	Two rods of lengths a and b slide along the axes in such a way that their ends are concyclic. The locus of the centre of the circle passing through these points is
Answer» `4(x^(2)+y^(2))=a^(2)+b^(2)` `x^(2)-y^(2)=a^(2)-b^(2)` `4(x^(2)-y^(2))=a^(2)-b^(2)` `x^(2)-y^(2)=a^(2)+b^(2)` Solution :Let `A_(1)A_(2) and B_(1)B_(2)` be two rods of length a and b which slide LONG OX and OY respectively. Let `x^(2) + y^(2) +2gx +2fy +c= 0` be the circle passing through `A_(1), A_(2), B_(1), B_(2)`.Then `A_(1)A_(2)`= Intercept of x-axis `=2 sqrt(g^(2)-c)` `rArr a=2 sqrt(g^(2)-c)`...(i) `B_(1)B_(2)`= intercept of y-axis `2sqrt(f^(2)-c)`...(ii) Squaring (i) and (ii), we get `a^(2)=4(g^(2)-c)` `b^(2)=4 (f^(2)-c)` Substracting `a^(2)-b^(2)=4(g^(2)-f^(2))` HENCE locus of CENTRE is `a^(2)-b^(2)=4(x^(2)-y^(2))`