If the chord y=mx+1 of the circle x^2+y^2=1 s | vedclass

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(C) The angle subtended by a chord at the center is twice the angle subtended by it at the major segment. Given the angle at the major segment is $45^

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) The angle subtended by a chord at the center is twice the angle subtended by it at the major segment. Given the angle at the major segment is $45^{\circ}$,the angle subtended at the center $C(0,0)$ is $2 	imes 45^{\circ} = 90^{\circ}$. Let $CP$ be the perpendicular distance from the center $(0,0)$ to the chord $mx-y+1=0$. $CP = \frac{|m(0) - (0) + 1|}{\sqrt{m^2 + (-1)^2}} = \frac{1}{\sqrt{m^2+1}}$. In the right-angled triangle formed by the center,the midpoint of the chord,and an endpoint of the chord,the angle at the center is half of $90^{\circ}$,which is $45^{\circ}$. Thus,$\cos 45^{\circ} = \frac{CP}{	ext{radius}} = \frac{CP}{1}$. $\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{m^2+1}}$. Squaring both sides,$m^2+1 = 2$,so $m^2 = 1$,which gives $m = \pm 1$. Since the options provided include $-1$ but not $1$,the correct choice is $-1$.

If the chord $y=mx+1$ of the circle $x^2+y^2=1$ subtends an angle of $45^{\circ}$ at the major segment of the circle,then the value of $m$ is:

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