Distance from a point to a set

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SUMMARY

The discussion focuses on calculating the distance from the point (4,2) to the set A2, defined as the unit circle with the equation x² + y² = 1. The distance is determined as the greatest lower bound of the distances from (4,2) to points (x,y) on the circle, expressed mathematically as ρ((4,2), (x,y)) = √(21 - 8x - 4y). Participants suggest two methods for finding the minimum distance: differentiating the distance function and using geometric principles to identify the intersection of the line through (0,0) and (4,2) with the circle.

PREREQUISITES
  • Understanding of Euclidean distance in a Cartesian plane
  • Knowledge of calculus, specifically differentiation
  • Familiarity with the equation of a circle
  • Basic geometry concepts related to lines and intersections
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  • Learn how to differentiate functions to find critical points
  • Study the geometric properties of circles and their intersections with lines
  • Explore optimization techniques in calculus for finding minimum distances
  • Investigate the concept of greatest lower bounds in mathematical analysis
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Homework Statement


I need to find [tex]\rho((4,2), A_2)[/tex] where [tex]A_2=\{(x,y)|x^2+y^2=1 \}[/tex].

Homework Equations





The Attempt at a Solution


I know that the definition says that this distance is the greatest lower bound of [tex]\{\rho(x,a)|a \in A\}[/tex]. So the distance between [tex]\rho((4,2), (x,y))= \sqrt{(x-4)^2+(y-2)^2}=\sqrt{x^2+8x+16+y^2-4y+4}= \sqrt{21-8x-4y}[/tex] and then I am not sure what to do.
 
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tarheelborn said:

Homework Statement


I need to find [tex]\rho((4,2), A_2)[/tex] where [tex]A_2=\{(x,y)|x^2+y^2=1 \}[/tex].

Homework Equations




The Attempt at a Solution


I know that the definition says that this distance is the greatest lower bound of [tex]\{\rho(x,a)|a \in A\}[/tex]. So the distance between [tex]\rho((4,2), (x,y))= \sqrt{(x-4)^2+(y-2)^2}=\sqrt{x^2+8x+16+y^2-4y+4}= \sqrt{21-8x-4y}[/tex] and then I am not sure what to do.
Well, that is [itex]D= (x^2+ y^2+ 8x- 4y+ 20)^{1/2}[/itex]. Differentiate that and set the derivative equal to 0 to find the minimum.

Or, use geometry! [itex]x^2+ y^2= 1[/itex] is a circle with center at the origin and radius 1. The nearest point to (4, 2) on that circle lies on the straight line from (0, 0) to (4, 2) (an extended radius). The equation of the line through (0, 0) and (4, 2) is y= x/2. Where does that line intersect the circle [itex]x^2+ y^2= 1[/itex]
 

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