If $ heta$ is the angle between the tangents drawn from the point $(2,3)$ to the circle $x^2+y^2-6 x+4 y+12=0$, then $ heta=$

Question

If $	heta$ is the angle between the tangents drawn from the point $(2,3)$ to the circle $x^2+y^2-6 x+4 y+12=0$, then $	heta=$

Aicharya · Accepted Answer

Given the equation of circle $\begin{aligned}& x^2+y^2-6 x+4 y+12=0 \\& \Rightarrow(x-3)^2+(y+2)^2=1\end{aligned}$

Since, $A O=\sqrt{(3-2)^2+(-2-3)^2}=\sqrt{1+25}=\sqrt{26}$ and, $A P=\sqrt{26-1}=5$ Now, in $\triangle \mathrm{AOP}, \tan \theta=\frac{1}{5} \Rightarrow \theta=\tan ^{-1}\left(\frac{1}{5}\right)$ Required angle $=2 \theta=2 \tan ^{-1}\left(\frac{1}{5}\right)$ $=\tan ^{-1}\left(\frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^2}\right)=\tan ^{-1}\left(\frac{10}{24}\right)=\tan ^{-1}\left(\frac{5}{12}\right)$

If $\theta$ is the angle between the tangents drawn from the point $(2,3)$ to the circle $x^2+y^2-6 x+4…

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