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10-1.Circle and System of Circles
hard
The centre of the circle passing through the point $(0,1)$ and touching the parabola $y=x^{2}$ at the point $(2,4)$ is
A
$\left(\frac{3}{10}, \frac{16}{5}\right)$
B
$\left(\frac{-16}{5}, \frac{53}{10}\right)$
C
$\left(\frac{6}{5}, \frac{53}{10}\right)$
D
$\left(\frac{-53}{10}, \frac{16}{5}\right)$
(JEE MAIN-2020)
Solution
$y=x^{2}$
$\left.\frac{d y}{d x}\right|_{P}=4$
$(y-4)=4(x-2)$
$4 x-y-4=0$
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 $:\left(2-2 \lambda, \frac{\lambda+8}{2}\right) \equiv\left(\frac{-16}{5}, \frac{53}{10}\right)$
Standard 11
Mathematics
