IIT-JEE 1983Question 1
The centre of the circle passing through the point $(0,1)$ and touching the curve $y=x^{2}$ at $(2,4)$ is
Circle
Solution
The centre lies on the normal at $(2,4)$: $(2-4t,4+t)$. Equal distances to $(2,4)$ and $(0,1)$: $17t^2=(2-4t)^2+(3+t)^2\Rightarrow t=\frac{13}{10}$, centre $\left(-\frac{16}{5},\frac{53}{10}\right)$.