The locus of the centres of the circles for which one end of a diameter is (1, 1) while the other end is on the line $$x + y = 3$$   is :

Solution :
The other end is $$\left( {t,\,3 - t} \right)$$
So the equation of the variable circle is
$$\eqalign{ & \left( {x - 1} \right)\left( {x - t} \right) + \left( {y - 1} \right)\left( {y - 3 + t} \right) = 0 \cr & {\text{or }}{x^2} + {y^2} - \left( {1 + t} \right)x - \left( {4 - t} \right)y + 3 = 0 \cr} $$
$$\therefore $$  the centre $$\left( {\alpha ,\,\beta } \right)$$  is given by $$\alpha = \frac{{1 + t}}{2},\,\,\beta = \frac{{4 - t}}{2}$$
$$ \Rightarrow \,\,2\alpha + 2\beta = 5$$