For each parabola y=x2+px+q, meeting coordinate axes at 3-distinct points, if circles are drawn through these points, then the family of circles must pass through
Suppose the parabola y=x2+px+q cuts X-axis at A(α,0)and B(β,0). Then, α,β are roots of the equation x2+px+q=0 ∴α+β=−p‌and‌α‌β=q The parabola y=x2+px+q cuts Y-axis at (0,q). Let the equation of the circle passing through A,B and C be x2+y2+2gx+2fy+c=0.....(i) ∴α2+2gα+c=0.......(ii) β2+2gβ+c=0....(iii) and q2+2fq+c=0...(iv) Subtracting Eq. (iii) from Eq. (ii), we get α+β+2g=0 ⇒g=
p
2
Adding Eqs. (ii) and (iii), we get α2+β2+2g(α+β)+2c=0 (α+β)2−2αβ+2g(α+β)+2c=0 p2−2q−p2+2c=0[∵α+β=pandg=
p
2
] Putting c=q in Eq. (iv), we get f=−(
q+1
2
) 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 x2+y2+px−(q+1)y+q=0 Clearly, it passes through (0,1).