Show that a parabola passes through the point (0,1)

Question

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 $(0,1)$.

Answer 1

Let $A$, $B$ be the points of intersection of the parabola with the $x$ axis , $C$ the point of intersection with the $y$ axis, $D$ the point of intersection of the circumcircle of $\triangle ABC$ with the $y$ axis and $O$ the origin. Since $ACBD$ is concyclic, $\triangle AOC \sim \triangle DOB$, so $OD = \dfrac{AO \cdot OB}{OC}$. Now looking at the equation $x^2 + ax + b$, $AO \cdot OB$ is the product of the roots and $OC = b$, so by vieta formulas $OD = 1$ as desired.

Answer 2

You may find the three distinct points first by solving quadratic equations. Then there is a circle passing through the three points, namely, $x^2 + y^2 + Dx + Ey + F = 0$. By substituting the three points into the circle, $D$, $E$ and $F$ can be solved by linear equations. Then you may check whether the point $(0,1)$ lies on the circle.

Answer 3

The points are
$$\{(x_1\mid y_1);(x_2\mid y_2);(x_3\mid y_3)$$
If
$$\begin{vmatrix} x_1^2+y_1^2&x_1&y_1&1\\ x_2^2+y_2^2&x_2&y_2&1\\ x_3^2+y_3^2&x_3&y_3&1\\ 1&0&1&1 \end{vmatrix}=0\;$$,
then $(0\mid 1)\;$ is concyclic with the other points.
For your parabola, the pertinent points are
$$\{\left(\frac{-a+\sqrt{a^2-4b}}2\mid 0\right);\left(\frac{-a-\sqrt{a^2-4b}}2\mid 0\right);(0\mid b)\}$$ and the corresponding determinant is

$$\begin{vmatrix} \frac{2a^2-4b-2a\sqrt{a^2-4b}}4&\frac{-a+\sqrt{a^2-4b}}2&0&1\\ \frac{2a^2-4b+2a\sqrt{a^2-4b}}4&\frac{-a-\sqrt{a^2-4b}}2&0&1\\ b^2&0&b&1\\ 1&0&1&1 \end{vmatrix}\;$$.

No matter what the values of $a$ and $b$ -- even if the parabola does not intersect with the x-axis -- the determinant always vanishes! So when the parabola DOES intersect the x-axis, the points are concyclic.