The angle between the two tangents drawn from origin to the circle $x^2+y^2-14 x+2 y+25=0$ is

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Aicharya · Accepted Answer

Given equation of circle is $ \begin{aligned} x^2+y^2-14 x+2 y+25 & =0 \\ \Rightarrow\left(x^2-14 x+49\right)+\left(y^2+2 y+1\right)-49-1+25 & =0 \\ (x-7)^2+(y+1)^2 & =25=5^2 \end{aligned} $

$\Rightarrow \quad$ Radius of circle $=5$ Centre of circle $=(7,-1)$ Distance between origin and centre $ \begin{aligned} & =\sqrt{(7)^2+(-1)^2}=\sqrt{50}=5 \sqrt{2} \\ \therefore \quad O P & =5 \sqrt{2} \text { and } P A=5 \end{aligned} $ Now, $\sin \theta=\frac{A P}{O P}=\frac{5}{5 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}$ Similarly, $\sin \alpha=\frac{P B}{O P}=\frac{5}{5 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \alpha=45^{\circ}$ $ \therefore \quad \theta+\alpha=90^{\circ} $ Thus, angle between two tangent drawn from origin to given circle is $90^{\circ}$

The angle between the two tangents drawn from origin to the circle $x^2+y^2-14 x+2 y+25=0$ is

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