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 other end be (t,-3,t)
So, the equation of the variable 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`
`:.` The centre, `(alpha, beta)` is given by
`alpha=(1+t)/(2),beta=(4-t)/(2)`
`rArr 2 alpha+ 2 beta=5`
Hence, the locus is `2x+2y=5`