When Nine Points Are Worth But Eight: Euler’s Resolution of Cramer’s Paradox - Conic Exceptions

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 $x=0;$ $y=0;$ $x=1;$ $y=1;$ $x=2;$ $y=2;$ $x=3;$ $y=3;$ $x=4;$ $y=4;$
all of the coefficients of the general equation $\alpha yy + \beta xy + \gamma xx + \delta y + \varepsilon x + \zeta = 0$ will not be determined, for after having introduced all of the given determinations, we are brought to this equation $\alpha yy - (\alpha + \gamma)xy + \gamma xx + \delta y - \delta x = 0,$ so that there still remain two coefficents to be determined. If from the five given points there had been but 4 arranged in a straight line, then there would remain but one coefficient to be determined.
Euler did not use ordered pairs as we do today, but he was asking Cramer to consider the problem of trying to fit the points $(0,0),$ $(1,1),$ $(2,2),$ $(3,3)$ and $(4,4)$ to the general equation of the second degree, $\alpha y^2 + \beta xy + \gamma x^2 + \delta y + \varepsilon x + \zeta = 0.$ (It's interesting to note that Euler wrote $yy$ and $xx$ where we would write $y^2$ and $x^2.$) In this case, it turns out there is no unique solution to this equation.