Distance from a point to a set

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tarheelborn
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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]