First, let's draw the graph and see what the question means.

*y* = 2*x*^{3} −* x* + 3

I have used **equal scaling** along the 2 axes (so that later, when I draw the circle, it will not have an elliptical shape).

Now, to find the radius of curvature, we need:

`(dy)/(dx)=6x^2-1`

And then

`((dy)/(dx))^2=(6x^2-1)^2`

`=36x^4-12x^2+1`

So now we are ready to substitute into the formula to give us the radius at **any** point *x*:

`R=[1+(dy/dx)^2]^(3"/"2)/(|(d^2y)/(dx^2)|)`

`=[1+36x^4-12x^2+1]^(3//2)/{|12x|}`

`=[36x^4-12x^2+2]^(3//2)/{|12x|}`

Now to find the radius of curvature at the required point *x* = 1, we substitute:

`[[36x^4-12x^2+2]^(3//2)/{|12x|}]_(x=1)` `=11.04787562`

To show what we have done, let's look at the graph of the curve (blue) with the approximating circle (dark red) overlaid. The circle is a good approximation for the curve at (1, 4).

We can show that the center of the approximating circle is `(−9.8, 6.17)`.

We know the length of the radius shown in the diagram (`11.05` units).

We know 1 point on that radius line, `(1,4)`, and we need to find the one at the other end, the center. Let's call it (*x*_{1}, *y*_{1}).

We're going to set up 2 equations using these 2 unknowns.

We know that

`(dy)/(dx)=6x^2-1`

So the slope of the tangent at `x = 1` is

6(1)

^{2}− 1 = 5.

Now, the slope of the **normal** (the line at right angles to the tangent at the point of contact) is −1/5. (See Tangents and Normals.)

So we can use the formula *y* − *y*_{1} = *m*(*x* − *x*_{1}) and the known slope `-1/5` and point `(1, 4)` to find the equation of the line containing the radius as follows:

`y = −x/5 + 21/5`

y− 4 = (−1/5) (x− 1), which gives:

The unknown point (*x*_{1}, *y*_{1}) lies on this line, so we can say

`y_1=-(x_1)/5+21/5`

Next, we use the distance formula:

`sqrt((x_2-x_1)^2+(y_2-y_1)^2)=r`

We know the distance *r* (since it is the radius), and one of our points is (1,4), and using our expression connecting *x*_{1} and *y*_{1}, we can solve these simultaneously as:

`sqrt((1-x_1)^2+[4-(21/5-1/5x_1)]^2)` `=11.04787562`

This is one equation in one unknown, which after some algebra gives us

x_{1}= − 9.833 (and another positive solution which doesn't apply here)

Substituting this back into `y_1=-(x_1)/5+21/5` gives

y_{1}= 6.167

So the center of the circle is `(−9.8, 6.17)`.