Let $y=x^2+ax+b$ cuts the coordinate axes at three distinct points. Show that the circle passing through these 3 points also passes through $(0,1)$. Let $y=x^2+ax+b$ be a parabola that cuts the coordinate axes at three distinct points. Show that the circle passing through these three points also passes through $(0,1)$.
Since, the graph of the function cuts the coordinate axes in 3 different points, the function has real and distinct solutions. The third point of intersection is at $(0,b)$. But, anything I do from here, looks like a dead end. Please help. Thank you.
 A: First of all, $b$ cannot be $0$. Why?
To make the explanation complete, I add $O = (0, 0)$ being the origin.
Case-1 $b \gt 0$ [a slightly difficult case and the analysis is left as an exercise.] 
Case-2 $b \lt 0$ 
As mentioned, the third point is $B(0, b)$. Note that B is on the negative y-axis and thus the length of $OB = – b$
As also mentioned, the function has two real and distinct roots (called $\alpha$, and $\beta$ with $\alpha \lt \beta$ ).
The product of roots = $\alpha \cdot \beta = b \lt 0$ implies that the two points on the x-axis are $H(\alpha , 0)$ and $K(\beta , 0)$ with $K$ on the positive axis and $H$ on the negative x-axis. Thus, the length of HO is $– \alpha$
Let the circle passing through B, H, K cuts the positive y-axis at $D(0, d)$ such that $d \gt 0$. [It must be. Draw a rough sketch to see why.]
Hope you have learnt “the power of a point”. Part of it says:- “If the diagonals of the cyclic quadrilateral PQRS intersect at X, then PX.XR = QX.XS.”
The points H, B, K, D together with O fit the above description. Therefore HO. OK = BO. OD
∴ $– \alpha \cdot \beta = – b \cdot d$
i.e. $– (b) = – b \cdot d$ 
With b cannot be $0$ in mind, the above yields $d = 1$.
A: Let the equation of circle is 
$x^2+y^2+2gx+2fy+c=0$
It should pass through $(0,b)$
Gives $\implies$ $b^2+2fb+c=0 $  [eq(1)}
Now the circle will intersect the $x$-axis at points given by the quadratic.
$x^2+0^2+2gx+2f(0)+c=0\implies x^2+2gx+c=0$ [eq(2)]
Now, as you know each quadratic has two roots $x_1 , x_2$. By setting $y=0$ , I am finding where this circle is intersecting with $x-axis$. And there are two points for it obviously. 
But these root must be the same as the roots formed by the intersection of parabola with $x-axis$ Because it is given that circle passes through the points where parabola cuts the axis. Hence parabola cuts the axis at?
$x^2+ax+b=0$[eq(3)]
Yet another quadratic equation with $2$ roots, but it is given that the points of intersection are common. Hence both the equation must be identical and representing two same roots. This is only possible if their coefficients are equal.
$2,3$ are identical.
$g=\frac{a}{2}$ , $b=c$
Now from (1) $b^2+2fb+b=0 \implies b(b+2f+1)=0\implies f=-\frac{1+b}{2}$, $b\neq 0$ (Degenerate case-neglected)
Hence the required circle. 
