Question

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

• A. \(2+\sqrt{2}\)
• B. \(-2+\sqrt{2}\)
• C. \(-1\pm\sqrt{2}\)
• D. \(\pm1\)

Hint:

For a chord subtending angle \(\theta\) at the center of a circle of radius \(r\): \[ \text{Chord length}=2r\sin\frac{\theta}{2} \] Also, distance of line \(Ax+By+C=0\) from origin is: \[ \frac{|C|}{\sqrt{A^2+B^2}} \]

Answer 1

The Correct Option is D

Step 1: Note the circle and the chord.
The circle $x^2+y^2=1$ has centre $O(0,0)$ and radius $1$. The chord is $y=mx+1$, that is $mx-y+1=0$.
Step 2: Turn the inscribed angle into a central angle.
A chord that makes an angle of $45^\circ$ at the circle makes twice that, $90^\circ$, at the centre. This is the inscribed angle rule.
Step 3: Find the chord length.
A chord subtending angle $\theta$ at the centre has length $2r\sin\dfrac{\theta}{2}$. With $\theta=90^\circ$, \[ \text{length}=2(1)\sin45^\circ=\sqrt2. \]
Step 4: Link length to the distance $d$ from the centre.
Also length $=2\sqrt{r^2-d^2}$. So \[ \sqrt2=2\sqrt{1-d^2}\Rightarrow 2=4(1-d^2)\Rightarrow d^2=\frac12,\; d=\frac{1}{\sqrt2}. \]
Step 5: Write $d$ using the line formula.
The distance of $O(0,0)$ from $mx-y+1=0$ is $\dfrac{|1|}{\sqrt{m^2+1}}$. Set it equal to $\dfrac{1}{\sqrt2}$: \[ \frac{1}{\sqrt{m^2+1}}=\frac{1}{\sqrt2}. \]
Step 6: Solve for $m$.
Square: $m^2+1=2$, so $m^2=1$ and $m=\pm1$. \[ \boxed{\pm1} \]