Question:
The centre of the circle passing through the point $(0,1)$ and touching the parabola $\mathrm{y}=\mathrm{x}^{2}$ at the point $(2,4)$ is :
Correct Option: , 2
Solution:
$y=x^{2}$
$\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{P}}=4$
$(y-4)=4(x-2)$
$4 x-y-4=0$
Circle : $(x-2)^{2}+(y-4)^{2}+\lambda(4 x-y-4)=0$
passes through $(0,1)$
$4+9+\lambda(-5)=0 \Rightarrow \lambda=\frac{13}{5}$
Circle $: x^{2}+y^{2}+x(4 \lambda-4)+y(-\lambda-8)+(20-4 \lambda)=0$
Centre : $\left(2-2 \lambda, \frac{\lambda+8}{2}\right) \equiv\left(\frac{-16}{5}, \frac{53}{10}\right)$