Let the tangent to the circle C_{1}: x^{2}+y^ | vedclass

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(C) The equation of the tangent to $C_{1}: x^{2}+y^{2}=2$ at $M(-1, 1)$ is given by $x(-1) + y(1) = 2$,which simplifies to $-x + y = 2$,or $x - y + 2

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$\frac{1}{6}$

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(C) The equation of the tangent to $C_{1}: x^{2}+y^{2}=2$ at $M(-1, 1)$ is given by $x(-1) + y(1) = 2$,which simplifies to $-x + y = 2$,or $x - y + 2 = 0$.Let $O(3, 2)$ be the center of $C_{2}$ and $r = \sqrt{5}$ be its radius.The distance $d$ from $O(3, 2)$ to the line $x - y + 2 = 0$ is $d = \frac{|3 - 2 + 2|}{\sqrt{1^{2} + (-1)^{2}}} = \frac{3}{\sqrt{2}}$.Let $P$ be the midpoint of chord $AB$. In $\Delta OPA$,$OA = r = \sqrt{5}$ and $OP = d = \frac{3}{\sqrt{2}}$.Then $AP = \sqrt{OA^{2} - OP^{2}} = \sqrt{5 - \frac{9}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.In $\Delta OAN$,$\angle OAN = 90^{\circ}$. Let $\angle AON = 	heta$. Then $	an 	heta = \frac{AP}{OP} = \frac{1/\sqrt{2}}{3/\sqrt{2}} = \frac{1}{3}$.In $\Delta OAN$,$AN = OA 	an 	heta = \sqrt{5} \cdot \frac{1}{3} = \frac{\sqrt{5}}{3}$.Also,$ON = \sqrt{OA^{2} + AN^{2}} = \sqrt{5 + \frac{5}{9}} = \sqrt{\frac{50}{9}} = \frac{5\sqrt{2}}{3}$.The area of $\Delta ANB = \frac{1}{2} \cdot AB \cdot PN$. Since $PN = \frac{AN^{2}}{ON} = \frac{5/9}{5\sqrt{2}/3} = \frac{1}{3\sqrt{2}}$.Area $= \frac{1}{2} \cdot (2 \cdot AP) \cdot PN = AP \cdot PN = \frac{1}{\sqrt{2}} \cdot \frac{1}{3\sqrt{2}} = \frac{1}{6}$.

Let the tangent to the circle $C_{1}: x^{2}+y^{2}=2$ at the point $M(-1, 1)$ intersect the circle $C_{2}: (x-3)^{2}+(y-2)^{2}=5$ at two distinct points $A$ and $B$. If the tangents to $C_{2}$ at the points $A$ and $B$ intersect at $N$,then the area of the triangle $ANB$ is equal to

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