If the tangents drawn at the points O(0,0) an | vedclass

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(C) The equation of the circle is $x^{2}+y^{2}-2x-4y=0$. The center is $C(1, 2)$ and the radius $R = \sqrt{1^{2}+2^{2}} = \sqrt{5}$.The tangent at $O(

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$\frac{5+3\sqrt{5}}{2}$

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(C) The equation of the circle is $x^{2}+y^{2}-2x-4y=0$. The center is $C(1, 2)$ and the radius $R = \sqrt{1^{2}+2^{2}} = \sqrt{5}$.The tangent at $O(0,0)$ is $x(0)+y(0)-(x+0)-2(y+0)=0$,which simplifies to $x+2y=0$.The tangent at $P(1+\sqrt{5}, 2)$ is $x(1+\sqrt{5})+y(2)-(x+1+\sqrt{5})-2(y+2)=0$,which simplifies to $x\sqrt{5} - 5 - \sqrt{5} = 0$,so $x = 1+\sqrt{5}$.Substituting $x = 1+\sqrt{5}$ into $x+2y=0$,we get $1+\sqrt{5}+2y=0$,so $y = -\frac{1+\sqrt{5}}{2}$.Thus,$Q = (1+\sqrt{5}, -\frac{1+\sqrt{5}}{2})$.The length of the tangent $L = OQ = \sqrt{(1+\sqrt{5}-0)^{2} + (-\frac{1+\sqrt{5}}{2}-0)^{2}} = \sqrt{(1+\sqrt{5})^{2} + \frac{(1+\sqrt{5})^{2}}{4}} = \sqrt{\frac{5(1+\sqrt{5})^{2}}{4}} = \frac{\sqrt{5}(1+\sqrt{5})}{2} = \frac{\sqrt{5}+5}{2}$.The area of $	riangle OPQ$ is given by $\frac{RL^{3}}{R^{2}+L^{2}}$.Substituting $R=\sqrt{5}$ and $L=\frac{5+\sqrt{5}}{2}$,the area simplifies to $\frac{5+3\sqrt{5}}{2}$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

If the tangents drawn at the points $O(0,0)$ and $P(1+\sqrt{5}, 2)$ on the circle $x^{2}+y^{2}-2x-4y=0$ intersect at the point $Q$,then the area of the triangle $OPQ$ is equal to

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