The centre of the circle that passes through the point $(0,1)$ and touches the curve $y=x^2$ at $(2,4)$ is
The centre of the circle that passes through the point $(0,1)$ and touches the curve $y=x^2$ at $(2,4)$ is
$\left(\frac{-16}{5}, \frac{27}{10}\right)$
$\left(\frac{-16}{7}, \frac{53}{10}\right)$
$\left(\frac{-16}{5}, \frac{53}{10}\right)$
$\left(\frac{-16}{5}, \frac{-53}{10}\right)$
Solution
Let equation of circle is $x^2+y^2+2 g x+2 f y+c=0$
Where centre is $(-\mathrm{g},-\mathrm{f})$
Since circle passes through $(0,1)$
$
\Rightarrow 0+1+0+2 \mathrm{f}+\mathrm{c}=0 \Rightarrow \mathrm{c}=2 \mathrm{f}-1
$
$(2,4)$ is also on the circle.
$
\begin{aligned}
& \therefore 4+16+4 \mathrm{~g}+8 \mathrm{f}-2 \mathrm{f}-1=0 \\
& \Rightarrow 4 \mathrm{~g}+6 \mathrm{f}-19=0
\end{aligned}
$
Only option (c) satisfy $4 g+6 f-19=0$