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

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Question

(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 at the center $C(0,0)$ is $2 	imes {45^\circ} = {90^\circ}$. Let $P$ be the foot of the perpendicular from the center $C(0,0)$ to the chord $y = mx + 1$. In the right-angled triangle $CPR$,where $CR$ is the radius $r = 1$,the angle $\angle PCR = {45^\circ}$. Thus,$CP = r \cos({45^\circ}) = 1 	imes \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$. The perpendicular distance from $(0,0)$ to $mx - y + 1 = 0$ is given by $\frac{|m(0) - 0 + 1|}{\sqrt{m^2 + (-1)^2}} = \frac{1}{\sqrt{m^2 + 1}}$. Equating the two expressions for $CP$: $\frac{1}{\sqrt{m^2 + 1}} = \frac{1}{\sqrt{2}}$ $m^2 + 1 = 2$ $m^2 = 1$ $m = \pm 1$. Since the options provided include $-1$ but not $1$,and the question asks for the value of $m$,the correct choice is $-1$.

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

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