Problem Analysis #7

We are given

$$f(x) = 2x^3 - 4x^2 + 3$$

and for $0 < t < 4$, $C_t$ is a circle through $(t, f(t))$ with the same tangent as $y = f(x)$ at that point, and its center is on the $x$-axis.

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Step 1 – Center and radius conditions

Let the center be $(c, 0)$ and radius $r$.
The circle passes through $(t, f(t))$, so

$$(t - c)^2 + (f(t))^2 = r^2. \tag{1}$$

The tangent to the circle at $(t, f(t))$ is perpendicular to the radius.
The slope of the radius from $(c, 0)$ to $(t, f(t))$ is

$$m_{\text{radius}} = \frac{f(t) - 0}{t - c}.$$

The slope of the tangent to $f$ at $t$ is $f'(t)$.
For the tangents to match, the radius must be perpendicular to the tangent line, so

$$\frac{f(t)}{t - c} \cdot f'(t) = -1. \tag{2}$$

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Step 2 – Find $c(t)$ and $r(t)$

From (2):

$$\frac{f(t)}{t - c} = -\frac{1}{f'(t)} \quad\Rightarrow\quad t - c = -f(t) f'(t).$$

Thus

$$c(t) = t + f(t) f'(t). \tag{3}$$

From (1):

$$r(t)^2 = (t - c)^2 + f(t)^2.$$

But $t - c = -f(t) f'(t)$, so

$$r(t)^2 = f(t)^2 f'(t)^2 + f(t)^2 = f(t)^2 \big( f'(t)^2 + 1 \big). \tag{4}$$

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Step 3 – Evaluate at $t = 1$

First,

$$f(1) = 2(1)^3 - 4(1)^2 + 3 = 2 - 4 + 3 = 1.$$

Derivative:

$$f'(x) = 6x^2 - 8x,$$

so

$$f'(1) = 6 - 8 = -2.$$

From (3):

$$c(1) = 1 + (1)(-2) = 1 - 2 = -1.$$

From (4):

$$r(1)^2 = (1)^2 \big( (-2)^2 + 1 \big) = 1 \cdot (4 + 1) = 5.$$

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Step 4 – Final sum

$$c(1) + r(1)^2 = -1 + 5 = 4.$$ $$\boxed{4}$$