The centre of the circle passing through the
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 :

  1. $\left(\frac{3}{10}, \frac{16}{5}\right)$

  2. $\left(\frac{-16}{5}, \frac{53}{10}\right)$

  3. $\left(\frac{6}{5}, \frac{53}{10}\right)$

  4. $\left(\frac{-53}{10}, \frac{16}{5}\right)$


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 : $\left(2-2 \lambda, \frac{\lambda+8}{2}\right) \equiv\left(\frac{-16}{5}, \frac{53}{10}\right)$

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