Question

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

• A. $\sqrt{2}$
• B. $1\pm \sqrt{2}$
• C. $-2\pm \sqrt{2}$
• D. $2\pm \sqrt{2}$

Answer 1

The correct option is B

$1\pm \sqrt{2}$

Equation of the chord is $y=mx+1$

Equation of the circle is $x^2+y^2=1$

The joint equation of the curve through the line and the circle can be given by $x^2+y^2=(y-mx{)}^2$

$=x^2(1-m^2)+2mxy=0$

Now, $\tan \theta =\pm \frac{2\sqrt{h^2-ab}}{a+b}$, where $a=1-m^2,b=0,$ and $h=m$

$\Rightarrow \tan \frac{\pi }{4}=\pm \frac{2\sqrt{m^2-0}}{1-m^2}$

$\Rightarrow 1-m^2=\pm 2m$

$\Rightarrow m^2\pm 2m-1=0$

$\Rightarrow m=\pm 1\pm \sqrt{2}$

$\Rightarrow m=1\pm \sqrt{2}$ and $-1\pm \sqrt{2}$