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(B) To find the angle subtended by the chord $x+y-1=0$ at the center of the circle $x^2+y^2=1$,we follow these steps:Step $1$: Identify the circle and

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$\frac{\pi}{2}$

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(B) To find the angle subtended by the chord $x+y-1=0$ at the center of the circle $x^2+y^2=1$,we follow these steps:Step $1$: Identify the circle and chord.The circle $x^2+y^2=1$ has its center at the origin $O(0,0)$ and radius $r=1$.The chord is given by the line $x+y-1=0$.Step $2$: Find the perpendicular distance $d$ from the origin to the chord.The distance $d$ from $(0,0)$ to $x+y-1=0$ is given by $d = \frac{|1(0)+1(0)-1|}{\sqrt{1^2+1^2}} = \frac{1}{\sqrt{2}}$.Step $3$: Calculate the angle.Let the chord intersect the circle at points $A$ and $B$. In the isosceles triangle $OAB$,the perpendicular from $O$ to $AB$ bisects the angle $\angle AOB$. Let $\angle AOB = 2	heta$.In the right-angled triangle formed by the origin,the midpoint of the chord,and one endpoint of the chord,we have $\cos(	heta) = \frac{d}{r} = \frac{1/\sqrt{2}}{1} = \frac{1}{\sqrt{2}}$.Thus,$	heta = 45^{\circ}$ or $\frac{\pi}{4}$ radians.The total angle subtended at the center is $\angle AOB = 2	heta = 2 	imes 45^{\circ} = 90^{\circ}$ or $\frac{\pi}{2}$ radians.

The angle subtended by the chord $x+y-1=0$ of the circle $x^2+y^2=1$ at the origin is

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