Ellipse and its eccentricity in terms of angle Φ 
The portion of the tangent to the ellipse ${x^2\over a^2}$ + ${y^2\over b^2}=1$  at the point $Φ$ is intercepted by the auxiliary circle subtends a right angle at the center of the ellipse. Find the eccentricity of the ellipse in terms of $Φ$.

Let $Q$ and $R$ be the points of intersection of the tangent at $P$ $(a\cos(Φ), b\sin(Φ))$ with the auxiliary circle
so I know that the equation of the auxiliary circle is
$$x^2 + y^2 = a^2 \tag{1}$$
and that the equation of the tangent $QR$ is
$$\frac{x}{a}\cos(Φ) + \frac{y}{b}\sin(Φ) = 1 \tag{2}$$
So now my book says that the combined equation of $CQ$ and $CR$ is determined by homogenizing equation $(1)$ with equation $(2)$, which gives
$$x^2 + y^2 = a^2 \left({x\over a}\cos(Φ) + {y\over b}\sin(Φ)\right)^2$$
What I concluded at first is maybe because equation $(2)$ gives a result of $1$ that it is no harm to multiply it by $a^2$ in equation $(1)$,  but why is equation $(2)$ squared in the combined equation? And what does that combined equation represent?
Thanks!
 A: If you have a set of points given by a system of equations, then every linear combination of those equations is also satisfied by that set of points.  
Applying this idea to the problem at hand, we know that the auxiliary circle $x^2+y^2=a^2$ and the tangent line $\frac xa\cos\phi+\frac yb\sin\phi=1$ both pass through $Q$ and $R$. We look for a linear combination of these equations that also passes through the origin. This will occur when the constant term of the combined equation is zero. We can find suitable coefficients by inspection: the first equation minus $a^2$ times the second will eliminat the constant term.  
The resulting equation described an ellipse, which isn’t particularly useful, so a bit of clever mathematical trickery is brought to bear: squaring both sides of the tangent line equation makes it a degenerate conic consisting of the tangent line and its reflection in the origin. By symmetry, the intersection of this additional line with the auxiliary circle are the reflections of $Q$ and $R$ and so lie on the same lines through the origin, so adding this extra line doesn’t really change the conditions of the problem. Subtracting $a^2$ times this squared equation from that of the auxiliary circle produces a degenerate conic: a pair of lines (perhaps coincident) that intersect at the origin. In fact they are precisely the lines through $Q$ and $R$. This is a pretty slick way to generate the required lines through the origin without explicitly computing $Q$ and $R$.  
From here, one could split the conic to get the individual equations of the lines and then apply the constraint that they must be perpendicular, but I suspect that the book takes a simpler approach. These lines are the asymptotes of a family of hyperbolas. Those asymptotes are perpendicular when the hyperbolas are rectangular, which in turn occurs when the sum of the coefficients of the squared terms (i.e., the trace of the matrix of the associated quadratic form) in the equation is zero.
