Let C be the circle x^2+y^2=1 in the XY-plane | vedclass

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(B) The equation of the circle $C$ is $x^2+y^2=1$. The line $L_t$ passes through $(0,1)$ and $(t,0)$. Its equation is $\frac{x}{t} + \frac{y}{1} = 1$,

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$\frac{\pi}{4}$

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(B) The equation of the circle $C$ is $x^2+y^2=1$. The line $L_t$ passes through $(0,1)$ and $(t,0)$. Its equation is $\frac{x}{t} + \frac{y}{1} = 1$,which simplifies to $y = 1 - \frac{x}{t}$. Substituting $y$ into the circle equation: $x^2 + (1 - \frac{x}{t})^2 = 1$. $x^2 + 1 - \frac{2x}{t} + \frac{x^2}{t^2} = 1$. $x^2(1 + \frac{1}{t^2}) = \frac{2x}{t}$. $x^2(\frac{t^2+1}{t^2}) = \frac{2x}{t} \implies x = 0$ or $x = \frac{2t}{1+t^2}$. The point $(0,1)$ corresponds to $x=0$. The other point $Q_t$ has $x = \frac{2t}{1+t^2}$. Then $y = 1 - \frac{2}{1+t^2} = \frac{1+t^2-2}{1+t^2} = \frac{t^2-1}{t^2+1}$. Let $t = 	an 	heta$. Then $x = \sin 2	heta$ and $y = -\cos 2	heta$. For $t \in [1, 1+\sqrt{2}]$,$	heta \in [\frac{\pi}{4}, \frac{3\pi}{8}]$. The angle $2	heta$ varies from $\frac{\pi}{2}$ to $\frac{3\pi}{4}$. The angle subtended by the arc at the origin is the difference in the polar angles of the endpoints: $\frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$.

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