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### Shortest Distance between a Point and a Circle

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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.

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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

-Doctor Charles,  The Math Forum

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Associated Topics:
High School Calculus

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