The angle between the tangents drawn from a p | vedclass

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(D) The equation of the circle is $x^2+y^2-2x-4y=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$,$f=-2$,and $c=0$. The center of the circle is

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$\frac{\pi}{2}$

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(D) The equation of the circle is $x^2+y^2-2x-4y=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$,$f=-2$,and $c=0$. The center of the circle is $C(-g, -f) = (1, 2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+(-2)^2-0} = \sqrt{1+4} = \sqrt{5}$. Let $P$ be the point $(4,3)$. The distance $d$ from $P$ to the center $C(1,2)$ is $d = \sqrt{(4-1)^2+(3-2)^2} = \sqrt{3^2+1^2} = \sqrt{9+1} = \sqrt{10}$. Let $	heta$ be the angle between the tangents. The angle between the radius and the tangent is $\alpha = \frac{	heta}{2}$. In the right-angled triangle formed by the center,the point $P$,and the point of tangency,we have $\sin(\alpha) = \frac{r}{d} = \frac{\sqrt{5}}{\sqrt{10}} = \frac{1}{\sqrt{2}}$. Thus,$\alpha = \frac{\pi}{4}$. The angle between the tangents is $	heta = 2\alpha = 2 	imes \frac{\pi}{4} = \frac{\pi}{2}$.

The angle between the tangents drawn from a point $(4,3)$ to the circle $x^2+y^2-2x-4y=0$ is

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