Official solution
Circle passes through A(0, 1) and B(2, 4).
y = x2
\[ \Rightarrow \] \[{\left. {{{dy} \over {dx}}} \right|_B}\] = 4
tangent at (2,4) is
(y -- 4) = 4(x -- 2)
4x -- y -- 4 = 0
Equation of circle
(x - 2)2
+ (y--4)2
+ \[\lambda (4x--y - 4) = 0
Passing through (0,1)
\[ \therefore \] 4 + 9 +
\[\lambda (--5) = 0
\[ \Rightarrow \] \[\lambda \] = \[{{13} \over 5}\]
\[ \therefore \] Circle is
x2-- 4x + 4 + y2
-- 8y + 16 +
\[{{13} \over 5}\][4x - y - 4] = 0
\[ \Rightarrow \] x2 + y2 + \[\left( {{{52} \over 5} - 4} \right)\]x - \[\left( {8 + {{13} \over 5}} \right)\]y + 20 - \[{{{52} \over 5}}\] = 0
\[ \Rightarrow \] x2 + y2 + \[{{{32} \over 5}x - {{53} \over 5}y}\] + \[{{48} \over 5}\] = 0
\[ \therefore \] Centre is \[\left( {{{ - 16} \over 5},{{53} \over {10}}} \right)\]