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Given $r < 57$, let $C$ denote a circle centered at ($57$, $r$) with radius $r$.

$C$ is tangent to a parabola, $y=x^2+r$, from the outside in the first quadrant.

Find the value of $r$.

I am not sure how to approach the above problem. I started with implicit differentiation of the general circle formula.

  1. $r^2 = (x-57)^2+(y - r)^2$

  2. $D(r^2) = D\left((x-57)^2+( y - r)\right)$

  3. $0 = 2x - 114 + y'(2y-2r)$

  4. $y' = 114-\dfrac{2x}{2y-2r}$

For parabola:

  1. $D(y)=D(x^2+r)$

  2. $y'=2x$

Which gives me the formula of the gradient of the tangent to the circle. However, I am unsure how to proceed from here. What does the question means by from the outside in the first quadrant?

I am actually modeled the parabola and circle in an online graph but I fail to see how does the first quadrant requirement comes into play since $r<57$.

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Let point of intersect be $(h,k)$. We get two equations, one from equating slopes and other from equating $y$ coordinate. $(h,k)$ satisfies both equations.

$$\frac{114-2h}{2k-2r} = 2h \tag{1}$$

$$(h-57)^2 + (k-r)^2 = r^2 \tag{2}$$

From first equation we can write:

$$114-2h = 4h^3$$

Since this is a cubic representing $x$ coordinate of point of intersection, we need only one real root. Which is the case here because derivative of cubic $4h^3+2h-114$ is positive for all real. The exact value might be cumbersome to calculate.

Now that we have $h$, use equation two to get $r$:

$$(h-57)^2 +h^4 = r^2$$

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The circle is tangent to the parabola at ($x_t$, $y_t$), so a vector $u_{\text{para}}$ collinear to the tangent of the parabola is perpendicular to the vector going from the tangential point to the circle center: $$v_{\text{circle}} = \begin{pmatrix} 57 - x_t\\ r - y_t \end{pmatrix} \; .$$

$u_{\text{para}}$ could be: $$u_{\text{para}} = \begin{pmatrix} 1\\ 2x_t \end{pmatrix} \; ,$$

We now calculate:

$$u_{\text{para}} \cdot v_{\text{circle}} = 0 \leftrightarrow 57 -x_t + 2 x_t (r -y_t) = 0 \; .$$

On the other end, the equation of the parabola let us simplify removing $r$ and $y_t$: $$ x_t^3 + \dfrac{1}{2} x_t - \dfrac{57}{2} = 0 \; .$$

This equation is the same @King Tut found. This kind of tricky equation with a cubic term can sometimes be solved easily. This one is part of them. Indeed, the family of cubic equations: $x^3+cx + d = 0$ can be solved using the Cardan's method which gives in the end : $$x = ^3\sqrt{\dfrac{-d + \sqrt{\Delta}}{2}}+^3\sqrt{\dfrac{-d - \sqrt{\Delta}}{2}} \; ,$$ where $\Delta = d^2 + \dfrac{4c^3}{27} \; .$

It gives us surprisingly: $x_t = 3 \; .$ Given that it is an exercice, I suppose you had to try the 5 first values by hand.

Later, we have thanks to the circle equation and the parabola equation: $$ r^2 = x_t^4 + (57 - x_t)^2$$

which gives approximately $r = 54.7449$ .

Here is the visual result in the end:

enter image description here

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