Q121 / 225 · JEE Main 2002 · Mathematics
If the chord $y=m x+1$ of the circle $x^2+y^2=1$ subtends an angle of measure $45^0$ at the major segment of the circle then value of $m$ is
- A$2 \pm \sqrt{2}$
- B$-2 \pm \sqrt{2}$
- Correct answerC$-1 \pm \sqrt{2}$
- Dnone of these
Solution
Equation of circles $x^2+y^2=1=(1)^2$
$$
\begin{aligned}
& \Rightarrow x^2+y^2=(y-m x)^2 \Rightarrow x^2=m^2 x^2-2 m x y \Rightarrow x^2\left(1-m^2\right)+2 m x y=0 \\
& \tan 45=\pm \frac{2 \sqrt{m^2-0}}{1-m^2}=\frac{\pm 2 m}{1-m^2} \Rightarrow 1-m^2=\pm 2 m \Rightarrow m^2 \pm 2 m-1=0 \\
& \Rightarrow m=\frac{-2 \pm \sqrt{4+4}}{2}=\frac{-2 \pm 2 \sqrt{2}}{2}=-1 \pm \sqrt{2}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.