Question

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

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

Answer 1

The correct option is C

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

Explanation for the correct option:

Finding the center of the circle:

The general equation of the circle is $x^2+y^2+2gx+2fy+c=0$.

The circle passes through $\left(0,1\right)$.

Substitute $x=0$ and $y=1$.

$$\begin{gathered} \Rightarrow 1+2f+c=0 \\ \Rightarrow c=-1-2f \end{gathered}$$

So, the equation become $x^2+y^2+2gx+2fy-2f-1=0$

Circle also passes through $\left(2,4\right)$

Substitute $x=2$ and $y=4$.

$$\begin{gathered} \Rightarrow 4+16+4g+8f-2f-1=0 \\ \Rightarrow 4g+6f+19=0 \end{gathered}$$

Now only $g=\frac{16}{5}$ and $f=\frac{-53}{10}$ satisfy this equation.

Therefore the correct answer is option (C).