Question

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

• A. $\left(\frac{16}{5},\frac{53}{10}\right)$
• B. $\left(\frac{-2}{3},\frac{-4}{3}\right)$
• C. $\left(\frac{-4}{3},\frac{2}{3}\right)$
• D. $\left(\frac{-16}{5},\frac{53}{10}\right)$

Answer 1

Answer: (D)

Equation of tangent to the curve $y=x^2$ at
$(2,4)$ is $\frac{y+4}{2}=2x$
$\Rightarrow 4x-y-4=0$
Equation of circle touching at $(2,4)$ and tangent $4x-y-4=0$ is
$(x-2{)}^2+(y-4{)}^2+\lambda (4x-y-4)=0\dots (i)$
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
$x^2-4x+4+y^2-8y+16+\frac{52}{5}x-\frac{13}{5}y-\frac{52}{5}=0$
$x^2+y^2-x\left(4-\frac{52}{5}\right)-y\left(8+\frac{13}{5}\right)+16-\frac{52}{5}=0$
Centre of circle $\left[\frac{1}{2}\left(4-\frac{52}{5}\right),\frac{1}{2}\left(8+\frac{13}{5}\right)\right]$
$=\left(\frac{-16}{5},\frac{53}{10}\right)$