Two rods of lengths `a` and `b` slide along the `x`-axis and `y`-axis respectively in such a manner that their ends are concyclic. The locus of the centre of the circle passing through the end points is:

Options

A. `4(x^(2)+y^(2))=a^(2)+b^(2)`
B. `4(x^(2)-y^(2))=a^(2)-b^(2)`
C. `x^(2)-y^(2)=a^(2)-b^(2)`
D. `x^(2)-y^(2)=4(a^(2)-b^(2))`

Answer

B

Explanation

Let `A_(1) A_(2)` and `B_(1)B_(2)` be two rods of lengths a and b which slide along OX and OY respectively. Let`x^(2)+y^(2)+2gx+2fy+c=0` be the circle passing through `A_(1), A_(2), B_(1)` and `B_(2)`. Then,
`A_(1)A_(2)`= Intercept on y-axis `= 2 sqrt(g^(2)-c)`
`rArr a=2 sqrt(g^(2)-c) " " ...(i)`
And,
`B_(1)B_(2)=` Intercept on y-axis `= 2sqrt(f^(2)-c)`
`rArr b=2 sqrt(f^(2)-c) " " ...(ii)`
Here, c is the variable which is to be eliminated to find the locus of the centre (-g, -f) of the circle `x^(2)+y^(2)+2gx+2fy+c=0`.
Eliminating c from (i) and (ii), we get `a^(2)-b^(2)=4(g^(2)-f^(2))`
Thus, the locus of (-g, -f) i.e. the centre of the circle is `a^(2)-b^(2)=4(x^(2)-y^(2))`