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  • The centre of the circle passing through and touching the curve <img alt="\mathrm{y=x^{2}}" src="https://entran

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

Option: 1

\left(-\frac{16}{5}, \frac{27}{10}\right)


Option: 2

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


Option: 3

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


Option: 4

 none of these


Answers (1)

best_answer

Equation of tangent to the curve \mathrm{y=x^{2}} at (2,4) is \mathrm{y-4 x+4=0} Circle passing through line \mathrm{y-4 x+4=0} is \mathrm{(\mathrm{x}-2)^{2}+(\mathrm{y}-4)^{2}+\lambda(\mathrm{y}-4 \mathrm{x}+4)=0}

Circle passing through (0,1) implies \lambda=-13 / 5

Equation of the required circle is
(\mathrm{x}-2)^{2}+(\mathrm{y}-4)^{2}-13 / 5(\mathrm{y}-4 \mathrm{x}+4)=0

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

Posted by

Kuldeep Maurya

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