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

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Aicharya · Accepted Answer

$(x+\lambda)^2+(y+1)^2=\lambda^2$

$\begin{aligned} & C \equiv(-\lambda,-1) r=\lambda \\ & \angle C O P=\frac{90^{\circ}}{2}=45^{\circ}\end{aligned}$ $\begin{aligned} \therefore \quad O P & =C P=\lambda \\ O C & =\sqrt{\lambda^2+1}\end{aligned}$ In $\triangle O C P,(O C)^2=(O P)^2+(P C)^2$ $\begin{aligned} & \Rightarrow \quad \lambda^2+1=\lambda^2+\lambda^2 \\ & \Rightarrow \quad \lambda^2=1\end{aligned}$

Suppose the angle between the tangents drawn from $(0,0)$ to the circle $(x+\lambda)^2+(y+1)^2=\lambda^2$ is…

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