If the chord $y=mx+1$ of the circle $x^2+y^2=1$ subtends an angle $\frac{\pi}{4}$ at the major segment of the circle, then find $m$

Question

By simple observations, it would be easy to deduce that $m=-1$

This can also be proved by the following figure

Where the line $y=-x+1$ subtends $\frac{\pi}{2}$ at the centre.

But the answer says $m=-1\pm \sqrt 2$

I found that the given value of m can be obtained by homogenising the equation, bu thay still doesn’t explain the diagram. Please clarify

Answer 1

Actually, the answer that I got was $\pm\left(1-\sqrt2\right)$.

The line $y=1+mx$ and the circle $x^2+y^2=1$ have two points in common: $(0,1)$ and $\left(-\frac{2m}{m^2+1},-\frac{m^2-1}{m^2+1}\right)$. You're after those $m$'s such that$$(0,1).\left(-\frac{2m}{m^2+1},-\frac{m^2-1}{m^2+1}\right)=\cos\left(\frac\pi4\right)=\frac1{\sqrt2}.$$This is equivalent to $\frac{1-m^2}{1+m^2}=\frac1{\sqrt2}$ and this occurs if and only if $m=\pm\left(1-\sqrt2\right)$.

Answer 2

Consider the lines $y=\pm x$;

1) $y=x$ intersects the circle in:

$2x^2=1$; $x=\pm 1/√2$; $y=\pm 1/√2$;

Discard negative $x$ solution (Why?).

Slope of line through $(0,1)$ and $(1/√2,1/√2)$ :

$m_1:=(1/√2-1)/(1/√2)=√2(1/√2-1)=1-√2.$

2) $y=-x$ intersects the circle in:

$2x^2=1$; $x=\pm 1/√2$; $y=\mp 1/√2$;

$x=-1/√2$; $y =1/√2$; discard negative $y$ solution.

Slope of line through $(0,1)$ and $(-1/√2,1/√2)$:

$m_2:=(1/√2-1)(-1/√2)=-(1-√2).$

Answer 3

The blue lines through $(0,1)$ are inclined at $\pm 22{\frac12}^{\circ}$ to horizontal with equations

$$ y= 1 \pm ( \sqrt{2} -1) x $$

each subtends $\pi/4$ at center.