If one end of the diameter is (1, 1) and the | vedclass

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(C) Let the coordinates of the other end of the diameter be $(t, 3-t)$ since it lies on the line $x+y=3$.Let the centre of the circle be $(h, k)$.The

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$2x+2y=5$

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(C) Let the coordinates of the other end of the diameter be $(t, 3-t)$ since it lies on the line $x+y=3$.Let the centre of the circle be $(h, k)$.The centre is the midpoint of the diameter with endpoints $(1, 1)$ and $(t, 3-t)$.So,$h = \frac{1+t}{2}$ and $k = \frac{1+3-t}{2} = \frac{4-t}{2}$.From these equations,we have $t = 2h-1$ and $t = 4-2k$.Equating the two expressions for $t$:$2h-1 = 4-2k$$2h+2k = 5$.Replacing $(h, k)$ with $(x, y)$,the locus of the centre is $2x+2y=5$.

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

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