The angle between the two tangents drawn from | vedclass

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(C) The given equation of the circle is $x^2+y^2-14x+2y+25=0$. Completing the square,we get $(x^2-14x+49) + (y^2+2y+1) - 49 - 1 + 25 = 0$,which simpli

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(C) The given equation of the circle is $x^2+y^2-14x+2y+25=0$. Completing the square,we get $(x^2-14x+49) + (y^2+2y+1) - 49 - 1 + 25 = 0$,which simplifies to $(x-7)^2 + (y+1)^2 = 25 = 5^2$. Thus,the radius $r = 5$ and the center $P = (7, -1)$. The distance from the origin $O(0,0)$ to the center $P(7,-1)$ is $OP = \sqrt{7^2 + (-1)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$. Let the tangents from the origin touch the circle at $A$ and $B$. In the right-angled triangle $	riangle OAP$,$\sin 	heta = \frac{AP}{OP} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$. Therefore,$	heta = 45^{\circ}$. Similarly,for $	riangle OBP$,$\sin \alpha = \frac{BP}{OP} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$,so $\alpha = 45^{\circ}$. The total angle between the tangents is $	heta + \alpha = 45^{\circ} + 45^{\circ} = 90^{\circ}$.

The angle between the two tangents drawn from the origin to the circle $x^2+y^2-14x+2y+25=0$ is (in $^{\circ}$)

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