What does it mean for the tangent to a circle from an interior point to be "imaginary"? My geometry text seems to say that the tangent to a circle from an interior point in the real plane is "imaginary".
Further... It seems that when a double cone is intersected by a plane with an angle greater than the angle of the generating line, it is said that "imaginary" lines pass through the vertex.
Basically, I just can't understand what is meant by "imaginary" tangents and lines. Pls shed light on the matter. :)
 A: When you try to find the intersection of two disjoint circles (or of a circle with a line that it doesn't intersect), when you solve the equations you will get a pair of conjugate complex solutions.
For example the intersection of the cirvle $x^2+y^2=1$ with the line $x=2$ gives you the complex points $(x=2,y=\pm i\sqrt 3)$. Of course you can't really plot them on your $\Bbb R^2$ sheet of paper.
Giving a geometric interpretation to those complex points (points of the complexified real plane) is a bit challenging.
For every real object (points, lines, and circles mainly) you can generalize them to complex objects by allowing complex coefficients.
Then between 2 points there always is a line, 2 lines are either parallel or secant, there always is a circle going through three points, two circles almost always intersect in two points, etc. 
In your case, the middle between the two tangency points is always a real point, and is also known as the image of the original point by the inversion around the circle. The inversion is involutive, so it gives an easy way to build it even if you start from inside the circle.

You can identify the plane $\Bbb R^2$ with the complex plane $\Bbb C$ with the map $(x,y) \mapsto (x+iy)$.
Now if you complexify that, you can extend the scalars by adding $j$ such that $j^2=-1$, and then you can identify $\Bbb C^2$ with $\Bbb C[j]/(j^2+1)$ with $(x=x_1+jx_2, y=y_1+jy_2) \mapsto (x+iy = x_1+jx_2+iy_1+ijy_2)$.
Now $\Bbb C[j]/(j^2+1)$, has a "natural" map into $\Bbb C^2$ defined by replacing $j$ with $i$ and $-i$ respectively. You end up with two elements of $\Bbb C$, so an ordered pair of points.
Concretely, from your point with complex coordinates $(x=x_1+jx_2, y=y_1+jy_2)$ you end up with the ordered pair of points $((x_1-y_2, y_1+x_2),(x_1+y_2, y_1-x_2))$ and this will have various properties.
A point is real when $x_2=y_2=0$, which is when the ordered pair is of the form (P,P) so you can see a very natural embedding of the real plane into our complexified plane (the set of ordered pair of points)
The complex conjugate of a point is obtained by changing the signs of $x_2$ and $y_2$, which transforms a pair $(P,Q)$ into the pair $(Q,P)$, so conjugation is easy to visualize, and a pair of conjugate complex points can be thought of as an unordered pair of points.
And you can see that so far, the complex structure of our complexified plane can be done in a "coordinate-free" way.
From two points you can substract one from the other to get a vector. This is just componentwise substraction : a complexified vector is a pair of vectors.
You can also rescale a vector. Scaling by a real number is the usual componentwise scaling (because after all, the important map is $\Bbb R$-linear).
The important thing is what happens when you multiply a vector by the complex unit $j$ : the first component of the vector gets rotated counterclockwise by $\pi/4$ and the second component is rotated just as much in the other direction.
Finally, you can get the "real part" of a complexified point by computing $(P+\bar P)/2$, and this corresponds to the real point which is the midpoint of the elements in the pair. More generally, you can do weighted means as long as as the sum of the coefficient is $1$, and this will have a geometrical meaning as in it will give you a complexified point independantly of what coordinates you choose. In contrast, the imaginary part of a complexified point (this would be $(P - \bar P)/2i$) doesn't have a geometrical meaning because the coefficients sum to $0$. Instead you should interpret it as the complexified vector between a complexified point and its real part.

Now, the usual real lines and real circles not only have real points on them, but they also have pairs of conjugate complexified points (that are not real). For real lines, they are the pairs that are symmetric around the real locus of the line. For real circles, they are the pairs where one point is the image of the other by the inversion around the circle.
(more generally, for almost all complex lines and complex circles, you go from the first component to the second by applying an antiholomorphic bijective rational map, but I won't describe what the nonreal lines and nonreal circles look like)
Usually you can construct the tangency points $Q,Q'$ from a point $P$ to a circle $C$ by drawing the circle $C'$ with diameter $[OP]$ and intersecting it with $C$.
When $P$ is inside the circle $C$, this circle $C'$ doesn't intersect $C$ at real points but at a pair of conjugate complexified points $(Q,Q')$ and $(Q',Q)$, For symmetry reasons, they are "located" somewhere on the line $(OP)$ and they are placed such that $Q'$ is the image of $Q$ by the inversions through both circles. 
Constructing them would need too much explanation, however we can determine their real part $R$ (which is the midpoint of $[QQ']$) : computation shows that $R$ is the image of $P$ by the inversion through $C$.
To construct $R$, draw $(OP)$, then draw the line perpendicular to $(OP)$ that goes through $P$, find its intersection points with $C$, draw the two tangents, they will intersect at $R$.
Since $Q$ and $Q'$ are "on $(OP)$" and symmetric with regards to $R$, they are complex points of the line normal to $(OP)$ going through $R$, so this can also reduce he problem to finding the complex intersections of a line with a circle.
