Mathematics - Circle Question with Solution | TestHub

Suppose the parabola $y=x^2+px+q$ cuts $X$-axis at $A(\alpha ,0)$ and $B(\beta ,0)$. Then, $\alpha ,\beta$ are roots of the equation $x^2+px+q=0$ $$\therefore \alpha +\beta =-p\text{ and }\alpha \beta =q$$ The parabola $y=x^2+px+q$ cuts $Y$-axis at $(0,q)$. Let the equation of the circle passing through $A,B$ and $C$ be $$\begin{array}{rlr}& x^2+y^2+2gx+2fy+c=0& ...(i)\\ \therefore & {\alpha }^2+2g\alpha +c=0& ...(ii)\\ & {\beta }^2+2g\beta +c=0& ...(iii)\\ \text{ and }& q^2+2fq+c=0& ...(iv)\end{array}$$ Subtracting Eq. (iii) from Eq. (ii), we get $$\begin{array}{rl}\alpha +\beta +2g& =0\\ \Rightarrow & g=p/2\end{array}$$ Adding Eqs. (ii) and (iii), we get $$\begin{array}{r}{\alpha }^2+{\beta }^2+2g(\alpha +\beta )+2c=0\\ (\alpha +\beta {)}^2-2\alpha \beta +2g(\alpha +\beta )+2c=0\\ p^2-2q-p^2+2c=0\end{array}$$ $$\left[\because \alpha +\beta =p\text{ and }g=\frac{p}{2}\right]$$ 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 $$x^2+y^2+px-(q+1)y+q=0$$ Clearly, it passes through $(0,1)$.