Question 1205247
<pre>
Ikleyn did the parabola problem for you.  I'll do the circle problem.</pre>1.Find the equation of the circle that passes through the points (0,0), (5,0),
and (3,3). Use both geometric and algebraic solution for comparison.<pre>OK, I'll try to show the straight-edge and compass geometric method to draw
a circle through three points.

{{{drawing(400,400,-6,6,-6,6, grid(1),circle(0,0,.12),circle(5,0,.12),circle(3,3,.12) )}}}
The geometric method is to find the center by:
1. drawing two chords,

{{{drawing(400,400,-6,6,-6,6, grid(1),circle(0,0,.12),circle(5,0,.12),circle(3,3,.12),line(0,0,3,3),
line(3,3,5,0) )}}}

2. drawing their perpendicular bisectors (in green):

{{{drawing(400,400,-6,6,-6,6, line(0,0,3,3),
line(3,3,5,0),grid(1),circle(0,0,.12),circle(5,0,.12),circle(3,3,.12),

green(line(-7,10,7,-4), line(-8,-6.5,7,3.5))


 )}}}
3. the center is where those two perpendicular bisectors of 2 chords intersect.
   (it looks like the point {{{(matrix(1,3,2&1/2,",",1/2))}}}, but we aren't sure
of it being exactly that with the geometric method.) 


4. Put the foot of the compass on the point where the perpendicular bisectors
   of the two chords intersect.  The open your compass to one of the points on
   the circle.  Then swing it around and draw a circle. 
{{{drawing(400,400,-6,6,-6,6, line(0,0,3,3),
line(3,3,5,0),grid(1),circle(0,0,.12),circle(5,0,.12),circle(3,3,.12),
circle(5/2,1/2,sqrt(13/2)),
green(line(-7,10,7,-4), line(-8,-6.5,7,3.5))


 )}}}

That's the geometric way to draw a circle through three points.

But the algebraic way is to begin with the general equation 
or the standard equation for a circle:'

It might be easier to substitute the three points in the standard equation:

{{{(x-h)^2+(y-k)^2=r^2}}}

{{{system((0-h)^2+(0-k)^2=r^2,(5-h)^2+(0-k)^2=r^2,(3-h)^2+(3-k)^2=r^2)}}}

{{{system(h^2+k^2=r^2,25-10h+h^2+k^2=r^2,9-6h+h^2+9-6k+k^2=r^2)}}}

Subtract the 1st equation from the other two equations:

{{{system(h^2+k^2=r^2,25-10h=0,9-6h+9-6k=0)}}}
{{{system(h^2+k^2=r^2,5-2h=0,18-6h-6k=0)}}}
{{{system(h^2+k^2=r^2,-2h=-5,3-h-k=0)}}}
Solve the 2nd equation for h
{{{system(h^2+k^2=r^2,h=5/2,3-h-k=0)}}}
Substitute 5/2 for h in the 3rd equation
{{{system(h^2+k^2=r^2,h=5/2,3-5/2-k=0)}}}
Multiply the 3rd equation by 2
{{{system(h^2+k^2=r^2,h=5/2,6-5-2k=0)}}}
{{{system(h^2+k^2=r^2,h=5/2,1-2k=0)}}}
{{{system(h^2+k^2=r^2,h=5/2,-2k=-1)}}}
{{{system(h^2+k^2=r^2,h=5/2,k=1/2)}}}
Substitute h=5/2, k=1/2 in the 1st equation:
{{{system((5/2)^2+(1/2)^2=r^2,h=5/2,k=1/2)}}}
{{{system(25/4+1/4=r^2,h=5/2,k=1/2)}}}
{{{system(26/4=r^2,h=5/2,k=1/2)}}}
{{{system(sqrt(13)/2=r,h=5/2,k=1/2)}}}

So the standard form is

{{{(x-5/2)^2+(y-1/2)^2=26/4}}}
{{{(x-5/2)^2+(y-1/2)^2=13/2}}}

So now we see that the center = (h,k) = {{{(matrix(1,3,5/2,",",1/2))}}}
and the radius is {{{sqrt(13)/2)}}}

So we graph the circle

{{{drawing(400,400,-6,6,-6,6, circle(5/2,1/2,.13),grid(1),circle(0,0,.12),circle(5,0,.12),circle(3,3,.12),

circle(5/2,1/2,sqrt(13/2))
)}}}

Other times we might just want the general equation

{{{x^2+y^2+ax+by+c = 0}}}

Substitute the three points in:
{{{system(0^2+0^2+a(0)+b(0)+c = 0,5^2+0^2+a(5)+b(0)+c = 0,3^2+3^2+a(3)+b(3)+c = 0)}}}
{{{system(c = 0,25+5a+c = 0,9+9+3a+3b+c = 0)}}}
{{{system(c = 0,25+5a+c = 0,18+3a+3b+c = 0)}}}
Substitute c=0 in the other two equations:
{{{system(c = 0,25+5a = 0,18+3a+3b = 0)}}}
Divide the 2nd equation by 5 and the third by 3
{{{system(c = 0,5+a = 0,6+a+b = 0)}}}
Solve the 2nd equation for 'a'
{{{system(c = 0,a = -5,6+a+b = 0)}}}
Substitute a = -5 in the third equation:
{{{system(c = 0,a = -5,6+(-5)+b = 0)}}}
Solve the 3rd equation for b
{{{system(c = 0,a = -5,b = -1)}}}

Substitute in the general solution:

{{{x^2+y^2+ax+by+c = 0}}}
{{{x^2+y^2-5x-1y+0 = 0}}}
{{{x^2+y^2-5x-y = 0}}}

But the trouble with the general equation of a circle is that we can't tell 
the center and radius from it unless we complete the square and get it into
the standard form.

The general form {{{x^2+y^2-5x-y = 0}}} and the standard form {{{(x-5/2)^2+(y-1/2)^2=13/2}}}
are equivalent.

Edwin</pre>