Question

Suppose a parabola $y=x^2-a x-1$ intersects the coordinate axes at three points A, B and C respectively. The circumcircle of $\triangle ABC$ intersects the $y$-axis again at the point $D (0, t )$. Find the value of $t$.

Answer 1

We have $x=\frac{a \pm \sqrt{a^2+4}}{2} $
$\alpha=\frac{a+\sqrt{a^2+4}}{2} ; \beta=\frac{a-\sqrt{a^2+4}}{2}$
Equation of family of circles through $A$ and $B$ is
$(x-\alpha)(x-\beta)+y^2+\lambda y=0$
As it passes through $C(0,-1)$,
$\alpha \beta+1-\lambda=0 $
$-1+1-\lambda=0 \Rightarrow \text { So } \lambda=0$
(But $\alpha \beta=-1)$
$\therefore$ Equation of circle through $A , B$ and $C$ is $( x -\alpha)( x -\beta)+ y ^2=0$
It cuts the $y$-axis when $x=0$, so $\alpha \beta+y^2=0 $ (Put $\left.\alpha \beta=-1\right)$
$y^2=1 \Rightarrow y=1 \text { or }-1$
Hence $t=1$