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(A) Let the centre of the circle be $(h, k)$. Since the centre lies on the line $x+y-5=0$,we have $h+k=5$,or $k=5-h$. Since the circle touches the lin

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$\pi$ sq. units

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(A) Let the centre of the circle be $(h, k)$. Since the centre lies on the line $x+y-5=0$,we have $h+k=5$,or $k=5-h$. Since the circle touches the lines $x=2$ and $y=5$,the radius $r$ of the circle is given by the distance from the centre $(h, k)$ to these lines. Thus,$r = |h-2|$ and $r = |k-5|$. Since the circle is in the first quadrant and touches $x=2$ and $y=5$,the centre must be to the right of $x=2$ and below $y=5$. So,$r = h-2$ and $r = 5-k$. Equating the two expressions for $r$: $h-2 = 5-k$,which implies $h+k=7$. However,we are given $h+k=5$. This suggests the circle must be in a position where the distances are $r = |h-2|$ and $r = |k-5|$. Given the centre $(h, k)$ lies on $x+y=5$ and the circle is in the first quadrant,let $r$ be the radius. The centre is $(2+r, 5-r)$ or $(2-r, 5+r)$ etc. Since it lies on $x+y=5$,$(2+r) + (5-r) = 7 
eq 5$. Let's re-evaluate: The distance from $(h, k)$ to $x=2$ is $r = |h-2|$ and to $y=5$ is $r = |k-5|$. Since the centre $(h, k)$ is in the first quadrant and on $x+y=5$,$h Thus $r = h-2$ and $r = 5-k$. Then $h = r+2$ and $k = 5-r$. Substituting into $h+k=5$: $(r+2) + (5-r) = 7$. This is a contradiction. Wait,if the circle touches $x=2$ and $y=5$,the centre is $(2+r, 5-r)$ or $(2-r, 5+r)$. If centre is $(2+r, 5-r)$,then $(2+r) + (5-r) = 7 
eq 5$. If centre is $(2+r, 5+r)$,then $(2+r) + (5+r) = 5$ $\Rightarrow 2r = -2$ $\Rightarrow r = -1$ (impossible). If centre is $(2-r, 5-r)$,then $(2-r) + (5-r) = 5$ $\Rightarrow 7-2r = 5$ $\Rightarrow 2r = 2$ $\Rightarrow r = 1$. For $r=1$,the centre is $(2-1, 5-1) = (1, 4)$. Check: $1+4=5$ (on the line). $r=1$. Area $= \pi r^2 = \pi(1)^2 = \pi$ sq. units.

Let the centre of the circle $S=0$ lie on the line $x+y-5=0$ and also lie in the first quadrant. If this circle touches both the lines $x-2=0$ and $y-5=0$,then the area of the circle is

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