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Equation of a Circle


Date: 08/25/97 at 19:59:33
From: anna slawson
Subject: Calculus

This is from Calculus, 4th edn, by Larson, Hostetler and Edwards.

Page 17, in the chapter on Cartesian Plane and Functions, Chap. 1.

Question:  Write the equation of the specified circle in general form.
 Points on Circle:  (1,-1) , (2,-2) , (0,-2).

The general form of the equation of the circle is:  
 Ax(squared) + Ay(squared) + Cx + Dy + F ===0, A does not equal 0.

I have tried all my tricks, to no avail.  Help! 


Date: 08/25/97 at 21:01:49
From: Doctor Scott
Subject: Re: Calculus

Hi Anna!

This is a great question!  I checked out my fifth edition of the 
Larson book you mentioned and it was there, too!  So, here's one 
approach.  I think there probably are others, but this one seems to 
work pretty well...

Remember the definition of a circle. It is the set of points 
equidistant from a fixed point, called the center.  This definition is 
important in solving this problem. 

Since we are talking about points being equidistant from a fixed 
point, and we are given the coordinates of the points on the circle, 
we should immediately think of the distance formula. Let's call the 
center of the circle (a, b). If we can find the coordinates for a 
and b, we know the center; from this we can find the radius, and we 
can write the standard form of a circle (x - a)^2 + (y - b)^2 = r^2.  
You can turn this into standard form by expanding the binomials.

If the center is (a, b) then we can write an expression for the radius 
of the circle from (a, b) to the three points given. These expressions 
must be equal to each other! In fact, since they are all equal, they 
are equal in pairs - so we have a system of equations we can solve for 
a and b. Here's the beginning of the work - you can finish it.

Let's label the given points: P(1, -1), Q(2, -2), R(0, -2), C(a, b)  
for convenience.

The distance PC is  SQRT[(b+1)^2 + (a-1)^2]
The distance QC is  SQRT[((b+2)^2 + (a-2)^2]
The distance RC is  SQRT[(b+2)^2 + (a-0)^2]

Now, PC = QC, so

(b+1)^2 + (a-1)^2 = (b+2)^2 + (a-2)^2   <-- I've squared both sides.

b^2 + 2b + 1 + a^2 - 2a + 1 = b^2 + 4b + 4 + a^2 - 4a + 4 <--expand

-6  =  2b - 2a.

So we have a relation between a and b.  Great!

Now, if we let QC = RC  (or PC = RC), we'll get another equation 
involving a and b.  I used PC = RC and got  2b + 2a = -2.

So, we can solve the two equations simultaneously to find a and b.  
This is the center of the circle. Then use the center to write the 
equation you need.

Good luck.  Hope this helped out!

-Doctor Scott,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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