Question:
The centre of the circle passing through the point $(0,1)$ and touching the parabola $y=x^{2}$ at the point $(2,4)$ is:
Correct Option: , 4
Solution:
(d) Circle passes through $A(0,1)$ and $B(2,4)$. So its centre is the point of intersection of perpendicular bisector of $A B$ and normal to the parabola at $(2,4)$.
Perpendicular bisector of $A B ;$
$y-\frac{5}{2}=-\frac{2}{3}(x-1) \Rightarrow 4 x+6 y=19$.......(i)
Equation of normal to the parabola at $(2,4)$ is,
$y-4=-\frac{1}{4}(x-2) \Rightarrow x+4 y=18$...(ii)
$\therefore$ From (i) and (ii), $x=-\frac{16}{5}, y=\frac{53}{10}$
$\therefore$ Centre of the circle is $\left(-\frac{16}{5}, \frac{53}{10}\right)$