Question

Suppose the angle between the tangents drawn from \((0,0)\) to the circle \[ (x+\lambda)^2+(y+1)^2=\lambda^2 \] is \(\frac{\pi}{2}\). Then \(\lambda\) satisfies:

• A. \(\lambda^2=1\)
• B. \(\lambda=0\)
• C. \(\lambda^2=4\)
• D. \(\lambda^2=9\)

Hint:

For tangents drawn from an external point to a circle, if the angle between tangents is \(\theta\), then use \[ \sin\frac{\theta}{2}=\frac{r}{d}, \] where \(r\) is radius and \(d\) is the distance of the external point from the centre.

Answer 1

The Correct Option is A

Step 1: Identify the centre and radius.
Given circle is \[ (x+\lambda)^2+(y+1)^2=\lambda^2. \] So, its centre is \[ C(-\lambda,-1) \] and radius is \[ r=|\lambda|. \] The external point is \[ O(0,0). \] Therefore, \[ OC=\sqrt{(-\lambda)^2+(-1)^2} = \sqrt{\lambda^2+1}. \]

Step 2: Use the angle between tangents formula.
If the angle between tangents from an external point is \(\theta\), then \[ \sin\frac{\theta}{2}=\frac{r}{OC}. \] Here, \[ \theta=\frac{\pi}{2}. \] So, \[ \sin\frac{\pi}{4} = \frac{|\lambda|}{\sqrt{\lambda^2+1}}. \]

Step 3: Substitute \(\sin\frac{\pi}{4}\).
\[ \frac{1}{\sqrt2} = \frac{|\lambda|}{\sqrt{\lambda^2+1}}. \] Squaring both sides, \[ \frac12=\frac{\lambda^2}{\lambda^2+1}. \] \[ \lambda^2+1=2\lambda^2. \] \[ \lambda^2=1. \]

Step 4: Final conclusion.
Therefore, \[ \boxed{\lambda^2=1} \]