If the circle (x-h)²+(y-k)²=r² touches the curve y=x²+1 at (1,2). Then show that h+2k=5.

Question

If the circle $$(x-h)^2+(y-k)^2=r^2\quad\quad\text{(a)}$$ touches the curve $$y=x^2+1\quad\quad\text{(b)}$$ at $(1,2)$. Then show that $h+2k=5$.

As Circle in equation (a) touches curve in equation (b), I tried to substitute $y=x^2+1$ in (a) and then putting $x=1$, I got $$h^2+k^2-2h-4k+5-r^2=0$$ which is apparently the general equation of circle. Then its center is $(1,2)$. And it satisfies $h+2k=5$. Please correct me.

Answer 1

I'm assuming by "touches" you mean "is tangent to," otherwise the relation is not necessary. Namely, we require the tangent line to each curve at $(1, 2)$ to be the same. For this, their slopes must be equal.

We find for the circle, using implicit differentiation,

\begin{align} 2(x - h) + 2(y_1 - k) \frac{dy_1}{dx} &= 0 \\ \frac{dy_1}{dx} &= -\frac{x - h}{y_1 - k}. \end{align}

For the parabola,

$$ \frac{dy_2}{dx} = 2x. $$

We require

$$ \frac{dy_1}{dx} \bigg|_{(1,2)} = \frac{dy_2}{dx} \bigg|_{(1,2)}. $$

Can you take it from here?

Answer 2

Show: $h + 2k = 5$.

The tangent to $y = x^2 + 1$ at $(1,2)$ is :

$y - 2 = 2 (x - 1)$ since $y'(1) = 2$.

The normal to this line passing through the same point $(1,2)$ is:

$y - 2 = -(1/2) (x -1)$ .

This line passes through the center $(h,k)$ of the given circle.

Hence:

$k - 2 = - (1/2) ( h - 1)$;

$- 2k +4 = h - 1$;

$h + 2k = 5$.

$\\$

P.S. Used:

1) Tangent to $y = x^2 + 1$ at $(1,2)$

has slope $m = y'(1) = 2$.

2) Line through $(1,2)$ with slope $m= 2$ is of the form:

$y - 2 = 2 ( x -1)$.

3)The normal to this line has slope$ -(1/m) = - (1/2)$.