Question

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. $(\frac{3}{10},\frac{16}{5})$
  
• B. $(\frac{6}{5},\frac{53}{10})$
  
• C. $(\frac{-16}{5},\frac{53}{10})$
  
• D. $(\frac{-53}{10},\frac{16}{5})$
  

Answer 1

The correct option is C $(\frac{-16}{5},\frac{53}{10})$


Circle passing through point $(0,1)$ and touching curve
$y=x^2$ at $(2,4)$
tangent at $(2,4)$ is
$\frac{(y+4)}{2}=x\left(2\right)$
$\Rightarrow y-4x+4=0$
Equation of circle
$(x-2{)}^2+(y-4{)}^2+\lambda (4x-y-4)=0$
Passing through $(0,1)$
$4+9+\lambda (-5)=0$
$\lambda =\frac{13}{5}$
Circle is
$x^2-4x+4+y^2-8y+16+\frac{13}{5}[4x-y-4]=0$
$x^2+y^2+(\frac{52}{5}-4)x-(8+\frac{13}{5})y+20-\frac{52}{5}=0$
$x^2+y^2+\frac{32}{5}x-\frac{53}{5}y+\frac{48}{5}=0$
Centre is $(-\frac{16}{5},\frac{53}{10})$