If the acute angle between the pair of tangents drawn from the origin to the circle $x^2+y^2-4 x-8 y+4=0$ is $\alpha$, then $ an \alpha=$

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If the acute angle between the pair of tangents drawn from the origin to the circle $x^2+y^2-4 x-8 y+4=0$ is $\alpha$, then $	an \alpha=$

Aicharya · Accepted Answer

$\begin{aligned} & 	ext {} \because \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-8 \mathrm{y}+4=0 \ & C \equiv \mathrm{x}^2+\mathrm{y}^2\left(-\mathrm{g}_1-\mathrm{f}ight)=(2,4) \mathrm{r}=\sqrt{4+16-4}=4\end{aligned}$ Equation of pair of tangents is $\mathrm{S} \mathrm{S}_1=\mathrm{T}^2$ Where : $S$ : Equation of circle $S_1=S(0,0)$ $ \begin{aligned} & \& \mathrm{~T}=\mathrm{xx}_1+\mathrm{y}_1+\mathrm{g}\left(\mathrm{x}+\mathrm{x}_1ight)+\mathrm{f}\left(\mathrm{y}+\mathrm{y}_1ight)+\mathrm{c}=0 \ & \Rightarrow\left(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-8 \mathrm{y}+4ight)(0+4)=(\mathrm{x} \cdot 0+\mathrm{y} \cdot 0-2(\mathrm{x}-0) \ & -4(\mathrm{y}-0)+4)^2 \ & \Rightarrow 4\left(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-8 \mathrm{y}+4ight)=(-2 \mathrm{x}-4 \mathrm{y}+4)^2 \ & \Rightarrow 12 \mathrm{y}^2+16 \mathrm{xy}=0 \end{aligned} $ Tangent of acute angle between pairs of straight line $a x^2+2 b x y+b y^2$ is $ 	an \alpha=\frac{2\left(h^2-a bight)}{a+b} $ In equation (i) $a=0, b=12, h=8$ $ 	herefore 	an \alpha=\frac{2\left(8^2-0ight)}{12}=\frac{4}{3} $

If the acute angle between the pair of tangents drawn from the origin to the circle $x^2+y^2-4 x-8 y+4=0$ is…

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