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(C) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since the circle touches $y=x^2$ at $(2,4)$,the normal to the parabola at $(2,4)$ mus

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since the circle touches $y=x^2$ at $(2,4)$,the normal to the parabola at $(2,4)$ must pass through the center $(h,k)$. The derivative of $y=x^2$ is $\frac{dy}{dx} = 2x$. At $x=2$,the slope is $4$. The slope of the normal is $-\frac{1}{4}$. The equation of the normal at $(2,4)$ is $y-4 = -\frac{1}{4}(x-2) \Rightarrow x+4y-18=0$. Thus,$h+4k=18$ (Equation $1$). The distance from $(h,k)$ to $(2,4)$ equals the distance from $(h,k)$ to $(0,1)$: $(h-2)^2 + (k-4)^2 = (h-0)^2 + (k-1)^2$. $h^2-4h+4 + k^2-8k+16 = h^2 + k^2-2k+1$. $-4h-6k+19=0 \Rightarrow 4h+6k=19$ (Equation $2$). Solving $h+4k=18$ and $4h+6k=19$: $h = 18-4k$. $4(18-4k)+6k=19$ $\Rightarrow 72-16k+6k=19$ $\Rightarrow 10k=53$ $\Rightarrow k=\frac{53}{10}$. $h = 18-4(\frac{53}{10}) = 18-\frac{106}{5} = \frac{90-106}{5} = -\frac{16}{5}$. The center is $(-\frac{16}{5}, \frac{53}{10})$.

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

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