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

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(C) Let the one end of the diameter be $A = (1, 1)$.Since the other end $B$ lies on the line $x + y = 3$,we can represent $B$ as $(t, 3 - t)$.The cent

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$2x + 2y = 5$

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(C) Let the one end of the diameter be $A = (1, 1)$.Since the other end $B$ lies on the line $x + y = 3$,we can represent $B$ as $(t, 3 - t)$.The centre $(h, k)$ of the circle is the midpoint of the diameter $AB$.Thus,$h = \frac{1 + t}{2}$ 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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