BITSAT 2022 Paper

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 $x^2+y^2+2gx +2fy+c=0 .....(i)$ $\therefore \alpha^2+2g\alpha+c=0 .......(ii)$ $\beta^2+2g\beta+c=0 ....(iii)$ and $q^2+2fq +c=0 ...(iv)$ Subtracting Eq. (iii) from Eq. (ii), we get $\alpha+\beta+2g=0$ $\Rightarrow g=\frac{p}{2}$ Adding Eqs. (ii) and (iii), we get $\alpha^2+\beta^2 +2g(\alpha+\beta)+2c=0$ $(\alpha+\beta)^2-2\alpha\beta+2g(\alpha+\beta)+2c=0$ $p^2-2q-p^2+2c=0 \left[\because \alpha+\beta=p \text{ and } g=\frac{p}{2}\right]$ 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 $(0, 1)$.