Question 1118854
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Circle at the origin, radius 5.  The radius to the given point, *[tex \LARGE (3,4)], must have the equation *[tex \LARGE y\ =\ \frac{4}{3}x].  Since the tangent to a circle at a point must be perpendicular to the radius at that point, the slope of the desired tangent must be *[tex \LARGE -\frac{3}{4}].  So use the point-slope form of an equation of a line to derive an equation for a line with the slope *[tex \LARGE -\frac{3}{4}] that passes through the point *[tex \LARGE (3,4)]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}

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