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Shortest Distance between a Point and a Circle


Date: 1/22/96 at 10:14:18
From: Anonymous
Subject: Calculus

Problem:  Find the shortest distance between the point (4,5) and
the circumference of x^2 + y^2 = 9.

The problem involves finding the derivative and then letting that
equal zero to find the minimum.  Where I'm stuck is I don't know 
how to set up the relationship between the circle and the point so 
that I get an equation to derive.

Please help!


Date: 5/30/96 at 14:54:23
From: Doctor Charles
Subject: Re: Calculus

Well you CAN use calculus.  Here's the technique:

Use polar coordinates to cut down x and y to one variable.  
Any point on the circle has coordinates (3 cos theta, 3 sin theta) for 
some theta. 

Then let L be the square of the distance of (4,5) from this point:

   L = (3*(cos theta) - 4)^2 + (3*(sin theta) - 5) ^2

Using (cos^2 theta + sin^2 theta = 1) you should get:

   L = 50 - 24 cos theta - 30 sin theta

Differentiate with respect to theta and get that tan theta = 4/5.  This 
will actually give you two values of theta.  One is the point on the 
circle furthest from (4,5) and the other gives you the one nearest (4,5). 
Substitute these into the formula of L to get the square of the 
distances.  Choose the smallest!

Without using calculus:

Note that (4,5) is sqrt(4^2 + 5^2) units away from the origin. The 
circle is the set of points 3 units from the origin.  If you draw a line 
from the origin to (4,5) then it will cut the circle at the point 
closest to (4,5). The length of the line from the circle to (4,5) must 
be:

        sqrt(4^2 + 5^2) - 3 = sqrt(41) - 3

The corresponding point on the circle has coordinates (a,b) given by:

        (a,b) = (3/sqrt(41)*4,3/sqrt(41)*5)
              = (12/sqrt(41) , 15/sqrt(41))

This is much simpler than the calculus method, although both will give 
the same answer.

-Doctor Charles,  The Math Forum

    
Associated Topics:
High School Calculus

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