Question

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

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

Answer 1

The correct option is C 2x + 2y = 5

The other end is (t, 3 - t)
So the equation of the variable circle is
(x - 1)(x - t) + (y - 1)(y - 3 + t) = 0
or $x^2+y^2-(1+t)x-(4-t)y+3=0$
$\therefore$ The centre $(\alpha ,\beta )$ is given by
$\alpha =\frac{1+t}{2},\beta =\frac{4-t}{2}$
$\Rightarrow 2\alpha +2\beta =5$
Hence, the locus is 2x + 2y = 5.