How would you explain this method of solving quadratic equations? I stumbled across this interesting geometrical method of solving quadratic equations. Can someone explain why are intersection points roots of equation? Why does circle have anything to do with quadratic equations? What is this method called?

 A: Here's the algebraic geometry behind it: for the quadratic $a x^2 + bx + c$, we have $AB = a$, $BC = b$, $CD = c$. Consider point $D$ to be at the origin.
The center of the circle is at $(b/2,(c-a)/2)$ and passes through the origin. This means its equation is
$$
\left(x-\frac{c-a}{2}\right)^2 + \left(y-\frac{b}{2}\right)^2 = \left(\frac{c-a}{2}\right)^2 + \left(\frac{b}{2}\right)^2.
$$
The circle intersects $BC$ at $x = c$. Plugging that in and rearranging gives
$$
\left(y-\frac{b}{2}\right)^2 = \left(\frac{c-a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 -\left(\frac{c+a}{2}\right)^2 = \frac{b^2-4ac}{4}
$$
At this point, you probably see where this is going:
$$
y = \frac{b\pm \sqrt{b^2-4ac}}{2},
$$
which is almost the quadratic formula. To get the actual roots, we want not the y coordinates, but the slope of the lines connecting $A$ and $x_1$ and $x_2$. This slope is
$$
m = \frac{-b\pm \sqrt{b^2-4ac}}{2a},
$$
which is of course the quadratic formula.
There is a generalization of this called Lill's method for solving higher-order polynomials.
Now, while this is the algebraic explanation, I'm not sure what the purely geometric argument is.
EDIT: In the event the quadratic has complex roots (such as $x^2 - 4x + 5$), you can use a circle construction to find them as follows:

$Bx/BA$ is the real part and $Ay/AB$ is the imaginary part. I leave the proof as an exercise for the reader.
A: First, let's label some points $BC$ intersects your circle in two points, let' call them  $E,F$, where $E$ is the one on the top and $F$ the on on the bottom. Let $G$ the intersection point between $DC$ and the circle.
Having that, note that $\angle AGD=90$ since $AD$ is diameter. Therefore, $AGCB$ is a rectangle, in particular $GC=1$ and even more, by symmetry you can see that $BE=FC=x_1$ and $BF=EC=x_2$ (I am just making labelings $x_1$,$x_2$ we will check that these  satisfy our quadratic equation in a moment.)
Note that $x_1+x_2=BE+EC=BC=4$.
Now, the power of $C$ with respect to the circle can be computed in two different ways: $(CG)(CD)=(CF)(CE)$, this is equivalent to $(1)(3)=x_1x_2$, i.e. $x_1x_2=3$.
So, $(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2=x^2-4x+3$, which implies that $x_1,x_2$ are roots of $x^2-4x+3$.
Note that this method works for any equation of the form $x^2-bx+c$.
A: 
$$\begin{align}
|BB'| = |PP'| \quad\to\quad b &= r+s \tag{1}\\
|PA||PB|=|PQ||PQ'| \quad\to\quad |PA||P'B'|=|PQ||P'Q|\quad\to\quad c &= rs \tag{2}
\end{align}$$
Therefore,
$$x^2 - b x + c = x^2 - (r+s)x+r s = ( x- r )( x- s) \tag{$\star$}$$
as one expects from Vieta's formulas.  $\square$

Note. Relation $(2)$ follows from computing the power of a point $P$ with respect to the circle. The reader should verify that everything works correctly when $A$ and $B'$ lie on opposite sides of $\overline{PP'}$ (corresponding to $c$ being negative). Also, the special case when $b=0$ is worth considering. (In particular, the figure for $x^2-c=0$ reverts to a classical construction of the geometric mean.)
A: 
I'll use this image from Wikipedia. Consider $f(x):=x^2-sx+p$. 
Observe that the circle has the following equation $$\big(x-\frac{s}2\big)^2+\big(y-\frac{p+1}2\big)^2=\frac{s^2+(p-1)^2}{4}$$
Letting $y=0$ and multiplying with $4$ $$\begin{align*}\big(2x-s\big)^2+p^2+2p+1=s^2+p^2-2p+1&\iff \big(2x-s\big)^2=s^2-4p\\&\iff x=\frac{s\pm\sqrt{s^2-4p}}2\end{align*}$$
which is the formula to evaluate the roots of $f$...
A: You can render the quadratic equation from this construction using nothing more sophisticated than the Pythagorean Theorem.
Begin by dropping the perpendicular from $A$ to $CD$, intersecting the latter line at $E$.  Thereby a right triangle $ADE$ and parallogram (rectangle) $ABCE$ formed, and applying the Pythagorean Theorem to the triangle (of course opposite sides of the rectangle are equal) then yields
$AD^2=BC^2+(CD-1)^2$
Now draw chords from $A$ and $D$ to either point of intersection, labeled $P$, of $\overline{BC}$ with the circle. Define $x$ as the distance from $B$ to this point of intersection.   Three more right triangles are formed, one of these having its right angle on the circle because this angle in inscribed in a semicircle.  Apply the Pythagorean Theorem to each:
$AP^2=x^2+1$
$DP^2=CD^2+(BC-x)^2$
$AD^2=AP^2+DP^2$
In the last of these equations substitute the right hand sides of the previous three equations for each term getting
$BC^2+(CD-1)^2=x^2+1+CD^2+(BC-x)^2$
$BC^2+CD^2-2CD+1=x^2+1+CD^2+BC^2-2CDx+x^2$
$-2CD=x^2-2BCx+x^2$
$\color{blue}{x^2-BCx+CD=0}$
The reader is encouraged to examine what happens if we make $\overline{CD}$ longer on the one hand, and if we reverse its direction on the other.
A: A quadratic equation can always be reduced to $x^2-bx+c=0$. Refer to the graph:
$\hspace{2cm}$
Note that $BE=x_1$ and $CE=x_2$, because from the Vieta $x_1+x_2=b$.
Note that $\Delta ABE$ and $\Delta CDE$ are similar:
$$\frac{AB}{CE}=\frac{BE}{CD} \iff \frac{1}{x_2}=\frac{x_1}{c} \iff x_1x_2=c,$$
which is the second part of the Vieta's formulas.
Similarly, one can get the relation for $BF=x_2$ using the similarity of $\Delta ABF$ and $\Delta CDF$.
A: 
The basic figure and method here appear early in Rene Descartes' Geometry (1637, tr. Smith & Latham, Dover, 1954). 
Even before beginning to develop what we now know as coordinate geometry, Descartes claims "it is possible to construct all the problems of ordinary geometry by doing no more than the little covered in the four figures I have explained" (p. 17).
The figure here is the fourth.  In the first two figures Descartes shows how to "multiply" and "divide" one line by another, and to take the square root of a line (the mean proportional between that line and an arbitrary unit line), thereby applying arithmetic/algebra to geometry. In the third he shows how to solve a quadratic equation when $b^2$ is preceded by a $+$ sign and $az$ by either $+$ or $-$.
Descartes did not generalize to a single quadratic equation and solution, but considered the possible cases separately: it took some time to get from Descartes to the "Cartesian coordinate system" as we know it.
As the translator's footnote 24 explains, the solution here rests on the Pythagorean theorem and Euclid Elements III, 36:$$MR\cdot MQ=(LM)^2$$ 
