D = (x - 4)^2 + (2x + 1 - 3)^2 = (x - 4)^2 + (2x - 2)^2.
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- oxed{ ext{A sphere of radius } \dfrac{3c}{2} ext{ centered at } \left(0, 0, \dfrac{3c}{2}
- Question: Find the point on the line \(y = 2x + 1\) in the plane that is closest to the point \((4, 3)\).
- Solution: We minimize the distance from a point \((x, 2x+1)\) on the line to \((4, 3)\). The squared distance is:
- D = (x^2 - 8x + 16) + (4x^2 - 8x + 4) = 5x^2 - 16x + 20.
- rac{dD}{dx} = 10x - 16 = 0 \Rightarrow x = rac{8}{5}.
- Then $y = 2\left( rac{8}{5}
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