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(C) The equation of the line at $(1,1)$ is $x+y-2=0$. The slope of this line is $-1$. Since the normal is perpendicular to the tangent,the slope of th

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$\sqrt{3}$

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(C) The equation of the line at $(1,1)$ is $x+y-2=0$. The slope of this line is $-1$. Since the normal is perpendicular to the tangent,the slope of the normal is $1$. Thus,$	an 	heta = 1$,which gives $	heta = \frac{\pi}{4}$. Let the centre of the circle be $(h, k)$. The coordinates of the centre are given by $h = 1 \pm r \cos 	heta$ and $k = 1 \pm r \sin 	heta$. Given $r = \sqrt{2}$ and $	heta = \frac{\pi}{4}$,we have: $h = 1 \pm \sqrt{2} \cos \frac{\pi}{4} = 1 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \pm 1$. $k = 1 \pm \sqrt{2} \sin \frac{\pi}{4} = 1 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \pm 1$. So,the possible centres are $(2, 2)$ or $(0, 0)$. The equations of the circles are $x^2 + y^2 = 2$ or $(x-2)^2 + (y-2)^2 = 2$. For the circle $x^2 + y^2 - 2 = 0$,the length of the tangent from $(1, 2)$ is $\sqrt{1^2 + 2^2 - 2} = \sqrt{1+4-2} = \sqrt{3}$. For the circle $(x-2)^2 + (y-2)^2 - 2 = 0$,the length of the tangent from $(1, 2)$ is $\sqrt{(1-2)^2 + (2-2)^2 - 2} = \sqrt{(-1)^2 + 0^2 - 2} = \sqrt{1-2} = \sqrt{-1}$,which is not possible. Thus,the length of the tangent is $\sqrt{3}$.

$A$ circle $S=0$ with radius $\sqrt{2}$ touches the line $x+y-2=0$ at $(1,1)$. Then,the length of the tangent drawn from the point $(1,2)$ to $S=0$ is

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