Question

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

• A. $(-\frac{16}{5},\frac{27}{10})$
• B. $(-\frac{16}{7},\frac{53}{10})$
• C. $(-\frac{16}{5},\frac{53}{10})$
• D. $(-\frac{16}{7},\frac{27}{10})$

Answer 1

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

gen. eqn of circle : $x^2+y^2+ax+by+c=0...\left(1\right)$

for pt (0,1) put it in gen eqn.

$1+b+c=0...\left(2\right)$

given that circle also passes (2,4) $\therefore$ put in gen eqn.

$4+16+2a+4b+c=0$

or $2a+4b+c=-20..\left(3\right)$

also $\because$ tangential to parabola $y=x^2,$ then slopes are same @ (2,4)

So $\frac{dy}{dx}=2x|=4$ (for parabola)

$x=2$

and for circle $2x+2y\frac{dy}{dx}+a+\frac{bdy}{dx}+0=0$

$x=2,\frac{dy}{dx}=4,y=4$

$4+8.4+a+4b=0$

or $a+4b=-36...\left(4\right)$

solving sys. of eqns (2), (3), (4)

$a=-36-4b$

$\downarrow$

$-72-8b+4b+c=-20$

$\Rightarrow -4b+c=52...\left(5\right)$

$b+c=-1$

or $-5b=53$ or $b=53/5$

and $c=\frac{48}{5}$

and $a=\frac{32}{5}$

then

${(x+\frac{a}{2})}^2+(y+\frac{b}{2}{)}^2$

$\frac{-a^2}{4}-\frac{b^2}{4}+c=0$

i.e centere of circle is at : $(\frac{-a}{2},\frac{-b}{2})$

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