API · no tools. This is the prompt and response recorded for this attempt; a full solution may not have been returned.
Return a single integer only. The problem has been revised so that no final answer is an irrational number, a parameterized expression, a range, a proof, a drawing, a graph, a locus equation, or a region description.
Let . For , let be the circle passing through , having the same tangent there as , and having its center on the -axis. If the center is and the radius is , find
f(x)=2x³-4x²+3, f'(x)=6x²-8x.
At t=1: f(1)=2-4+3=1, f'(1)=6-8=-2.
The circle has center (c,0) on x-axis. The radius to the point (1,1) must be perpendicular to the tangent line (slope -2). So the normal line at (1,1) has slope 1/2.
Center lies along the normal line from the point. Normal direction slope = 1/2. Line: y - 1 = (1/2)(x - 1). Set y=0: -1 = (1/2)(x-1), x-1 = -2, x = -1.
So c = -1.
r² = (1-(-1))² + (1-0)² = 4 + 1 = 5.
c + r² = -1 + 5 = 4.
\boxed{4}
Comparative Analysis (P7) (Average across runs)
Averages use reported measurements only. Missing usage is not treated as zero.