Problem Analysis #7

To solve the problem, we need to find the center $(c(t), 0)$ and radius $r(t)$ of the circle $C_t$ passing through $(t, f(t))$ with the same tangent as $y = f(x)$ at that point, and having its center on the $x$-axis.

Step 1: Find the derivative $f'(x)$ to determine the slope of the tangent line.

Given $f(x) = 2x^3 - 4x^2 + 3$, the derivative is:

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

At $x = t$, the slope of the tangent line is:

$$f'(t) = 6t^2 - 8t.$$

Step 2: Determine the equation of the tangent line at $(t, f(t))$.

The tangent line equation is:

$$y - f(t) = f'(t)(x - t).$$

Substituting $f(t) = 2t^3 - 4t^2 + 3$ and $f'(t) = 6t^2 - 8t$, we get:

$$y - (2t^3 - 4t^2 + 3) = (6t^2 - 8t)(x - t).$$

Step 3: Find the center $(c(t), 0)$ of the circle.

The center $(c(t), 0)$ lies on the normal line to the tangent line at $(t, f(t))$. The slope of the normal line is the negative reciprocal of the tangent slope:

$$\text{slope of normal} = -\frac{1}{f'(t)} = -\frac{1}{6t^2 - 8t}.$$

The equation of the normal line passing through $(t, f(t))$ is:

$$y - f(t) = -\frac{1}{6t^2 - 8t}(x - t).$$

Since the center $(c(t), 0)$ lies on this line, substituting $y = 0$ gives:

$$0 - f(t) = -\frac{1}{6t^2 - 8t}(c(t) - t).$$

Solving for $c(t)$:

$$- (2t^3 - 4t^2 + 3) = -\frac{c(t) - t}{6t^2 - 8t},$$ $$c(t) - t = (2t^3 - 4t^2 + 3)(6t^2 - 8t),$$ $$c(t) = t + (2t^3 - 4t^2 + 3)(6t^2 - 8t).$$

Expanding and simplifying:

$$c(t) = t + 12t^5 - 16t^4 + 18t^3 - 24t^3 + 32t^2 - 18t^2 + 24t,$$ $$c(t) = 12t^5 - 16t^4 - 6t^3 + 4t^2 + 25t.$$

Step 4: Find the radius $r(t)$.

The radius $r(t)$ is the distance from the center $(c(t), 0)$ to the point $(t, f(t))$:

$$r(t) = \sqrt{(c(t) - t)^2 + (f(t) - 0)^2}.$$

Since $c(t) - t = (2t^3 - 4t^2 + 3)(6t^2 - 8t)$, we have:

$$r(t)^2 = (2t^3 - 4t^2 + 3)^2(6t^2 - 8t)^2 + (2t^3 - 4t^2 + 3)^2.$$

Factoring out $(2t^3 - 4t^2 + 3)^2$:

$$r(t)^2 = (2t^3 - 4t^2 + 3)^2[(6t^2 - 8t)^2 + 1].$$

Step 5: Evaluate at $t = 1$.

Substitute $t = 1$ into $c(t)$ and $r(t)^2$:

$$c(1) = 12(1)^5 - 16(1)^4 - 6(1)^3 + 4(1)^2 + 25(1) = 12 - 16 - 6 + 4 + 25 = 21.$$ $$r(1)^2 = (2(1)^3 - 4(1)^2 + 3)^2[(6(1)^2 - 8(1))^2 + 1] = (2 - 4 + 3)^2[6 - 8]^2 + 1 = 1^2(-2)^2 + 1 = 4 + 1 = 5.$$

Final Answer:

$$c(1) + r(1)^2 = 21 + 5 = \boxed{26}.$$