For each parabola $y=x^2+p x+q$, meeting coordinate axes at 3-distinct points, if circles are drawn through…
For each parabola $y=x^2+p x+q$, meeting coordinate axes at 3-distinct points, if circles are drawn through these points, then the family of circles must pass through
$(1,0)$
$(0,1)$
$(1,1)$
$(p, q)$
Solution
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}
& x^2+y^2+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 } & q^2+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 \\
p^2-2 q-p^2+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 $(0,1)$