The angle between the tangents drawn from the | vedclass

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(A) The equation of the circle is $(x-7)^2+(y+1)^2=25$. The center of the circle is $C(7, -1)$ and the radius $r = 5$. The distance $d$ from the origi

Vedclass · Accepted Answer

$2 \arcsin(5/\sqrt{50})$

Vedclass · Answer

(A) The equation of the circle is $(x-7)^2+(y+1)^2=25$. The center of the circle is $C(7, -1)$ and the radius $r = 5$. The distance $d$ from the origin $O(0, 0)$ to the center $C(7, -1)$ is $d = \sqrt{7^2 + (-1)^2} = \sqrt{49+1} = \sqrt{50}$. Let $	heta$ be the angle between the tangents. The angle between the tangents is given by $2 \alpha$,where $\sin(\alpha) = \frac{r}{d}$. Thus,$\sin(\alpha) = \frac{5}{\sqrt{50}}$. Therefore,the angle between the tangents is $2 \arcsin\left(\frac{5}{\sqrt{50}}ight)$.

The angle between the tangents drawn from the origin to the circle $(x-7)^2+(y+1)^2=25$ is

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