Question 985482
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KMSTs solution is more general, and therefore better, IMHO, than what I am about to show you.  However, this is a slightly different way to look at this particular problem and may give you some additional insight.


Note that a tangent to a circle is perforce perpendicular to a radius.  An examination of the diagram shows us that a tangent as described, the perpendicular radius, and that segment of the *[tex \Large y]-axis between the origin and the *[tex \Large y]-intercept of the desired line form a right triangle.  Since the measure of the radius of the circle which forms the short leg of the right triangle is 2 and the measure of the *[tex \Large y]-axis segment that forms the hypotenuse of the right triangle measures 4, we have a 30-60-90 right triangle, where the angle between the radius and the *[tex \Large y] axis is the 60 degree angle.  See figure below:


 *[illustration tangentstocircle].


That means that the angle between the radius ray and the *[tex \Large x]-axis is 30 degrees.  A quick examination of a unit circle and using a multiplier of 2 to match the radius of the given circle, we see readily that the point of tangency in QI is *[tex \Large \left(\sqrt{3},1\right)].


Using the slope formula, the slope of the line that passes through the points *[tex \Large \left(\sqrt{3},1\right)] and *[tex \Large \left(0,4\right)] is *[tex \Large -\sqrt{3}].


Symmetry gives us a second point of tangency in QII, which is simply *[tex \Large \left(-\sqrt{3},1\right)] and then the slope of the line through the second point is just the additive inverse of the slope of the first equation.


From there it should be clear that any slope in the open interval *[tex \Large \left(-\sqrt{3},\sqrt{3}\right)] will produce a secant line.


Further, any slope in either open interval *[tex \Large \left(-\infty,-\sqrt{3}\right)] or *[tex \Large \left(\sqrt{3},\infty\right)] will produce a line that does not intersect the circle.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \