For each parabola y=x^2+p x+q , meeting coordinate axes at...

Suppose the parabola $y=x^2+p x+q$ cuts $X$-axis at $A(\alpha, 0)$ and $B(\beta, 0)$. Then, $\alpha, \beta$ are roots of the equation $x^2+p x+q=0 \therefore \quad \alpha+\beta=-$ p $\text{ and } \alpha \beta=$ q The parabola $y=x^2+p x+q$ cuts $Y$-axis at $(0, q)$. Let the equation of the circle passing through $A, B$ and $C$ be $$\begin{aligned}& x²+y²+2 g x+2 f y+c =0 &...(i)\\ \therefore \alpha^2+2 g \alpha+ c =0 &...(ii)\\ \beta^2+2 g \beta+ c =0 &...(iii)\\ \text{ and }& q²+2 f q+c =0 &...(iv) \end{aligned}$$Subtracting Eq. (iii) from Eq. (ii), we get $$\begin{array}{rlrl}\alpha+\beta+2 g & =0 \\ \Rightarrow & g = p / 2 \end{array}$$Adding Eqs. (ii) and (iii), we get $$\begin{array}{r}\alpha^2+\beta^2+2 g(\alpha+\beta)+2 c=0 \\ (\alpha+\beta)^2-2 \alpha \beta+2 g(\alpha+\beta)+2 c=0 \\ p²-2 q-p²+2 c=0 \end{array}$$ \left[\because \alpha+\beta=$p$\text{ and }$g=$\frac{p}{2}\right]$Putting$c=q$in Eq. (iv), we get$f=- \left(\frac{q+1}{2}\right) Substituting the values of $g, f$ and $c$ in Eq. (i), we obtain the equation of family of circles passing through $A, B$ and $C$ as x^2+y^2+p x-(q+1) y+q=0 Clearly, it passes through $(0,1)$.