Thank you for reporting, we will resolve it shortly
Q.
The centre of the circle passing through the point $(0,1)$ and touching the curve $y=x^{2}$ at $(2,4)$ is
TS EAMCET 2019
Solution:
Equation of tangent to the curve $y=x^{2}$ at
$(2,4)$ is $\frac{y+4}{2} =2 x $
$\Rightarrow \, 4 x-y-4=0$
Equation of circle touching at $(2,4)$ and tangent $4 x-y-4=0$ is
$(x-2)^{2}+(y-4)^{2}+\lambda(4 x-y-4)=0\,\,\,\,\,\,\dots(i)$
Since, this circle is also passes through $(0,1)$
$\therefore (0-2)^{2}+(1-4)^{2}+\lambda(0-1-4)=0$
$ 4+9-5 \lambda =0 $
$ \lambda =\frac{13}{5} $
Putting the value of $\lambda$ in Eq. (i), we get
$x^{2}-4 x+4+y^{2}-8 y+16+\frac{52}{5} x-\frac{13}{5} y-\frac{52}{5}=0$
$x^{2}+y^{2}-x\left(4-\frac{52}{5}\right)-y\left(8+\frac{13}{5}\right)+16-\frac{52}{5}=0$
Centre of circle $\left[\frac{1}{2}\left(4-\frac{52}{5}\right), \frac{1}{2}\left(8+\frac{13}{5}\right)\right]$
$=\left(\frac{-16}{5}, \frac{53}{10}\right)$