The centre of the circle passing through the | vedclass

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Question

$y=x^{2}$

$\left.\frac{d y}{d x}\right|_{P}=4$

$(y-4)=4(x-2)$

$4 x-y-4=0$

Circle $:(x-2)^{2}+(y-4)^{2}+\lambda(4 x-y-4)=0$

passes throu

Vedclass · Accepted Answer

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

Vedclass · Answer

$y=x^{2}$

$\left.\frac{d y}{d x}\right|_{P}=4$

$(y-4)=4(x-2)$

$4 x-y-4=0$

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 $:\left(2-2 \lambda, \frac{\lambda+8}{2}\right) \equiv\left(\frac{-16}{5}, \frac{53}{10}\right)$

The centre of the circle passing through the point $(0,1)$ and touching the parabola $y=x^{2}$ at the point $(2,4)$ is

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