If one end of the diameter is (1, 1) and the other end lies on the line `x+y=3` , then find the locus of the center of the circle.

Options

A. `x+y=1`
B. `2(x-y)=5`
C. `2x+2y=5`
D. none of these

Answer

C

Explanation

Let the coordinates of the other end of the diameter be (t, 3-t). Then, the equation of the circle is
`(x-1)(x-t)+(y-1)(y-3+t)=0`
`rArr x^(2)+y^(2)-(1+t)x-(4-t)y+3=0`
Let (h, k) be the coordinates of the centre.
`:. H=(1+t)/(2), k=(4-t)/(2)`
`rArr 2h-1=-2k+4 " " `[On eliminating t]
`rArr 2(h+k)=5`
Hence, the locus of (h, k) is `2(x+y)=5`.