Question 748607
the circle passes through the point (0,0), (5,0) and (3,3)
<pre>
You can do this either by using the general form or the standard form.
I'll first do it with the general form:

x² + y² + Dx + Ey + F = 0

Substitute (x,y) = (0,0)

0² + 0² + D(0) + E(0) + F = 0
                        F = 0
So we have

x² + y² + Dx + Ey + 0 = 0
    x² + y² + Dx + Ey = 0

Substitute (x,y) = (5,0)

5² + 0² + D(5) + E(0) = 0
              25 + 5D = 0
              25 + 5D = 0
                   5D = -25
                    D = -5

So we have:

    x² + y² + Dx + Ey = 0
    x² + y² - 5x + Ey = 0

Substitute (x,y) = (3,3)

        x² + y² - 5x + Ey = 0
(3)² + (3)² - 5(3) + E(3) = 0
         9 + 9 - 15 + 3E = 0
                  3 + 3E = 0
                      3E = -3
                       E = -1  

        x² + y² - 5x + Ey = 0
         x² + y² - 5x - y = 0

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You can also use the standard form equation for a circle:

      (x - h)² + (y - k)² = r²

Substitute (x,y) = (0,0)

      (0 - h)² + (0 - k)² = r²
                  h² + k² = r²

Substitute (x,y) = (5,0)

      (5 - h)² + (0 - k)² = r²
       25 - 10h + h² + k² = r²

Substitute (x,y) = (3,3)

      (3 - h)² + (3 - k)² = r²
9 - 6h + h² + 9 - 6k + k² = r²
   18 - 6h - 6k + h² + k² = r²

So we have this system of equations:

                  h² + k² = r²
       25 - 10h + h² + k² = r²
   18 - 6h - 6k + h² + k² = r² 

If we subtract the first equation term by term from each of the
other two equations we have:

       25 - 10h = 0
   18 - 6h - 6k = 0

Solving the first for h

       25 - 10h = 0
           -10h = -25 
              h = {{{(-25)/(-10)}}}
              h = {{{5/2}}}

Substituting in

   18 - 6h - 6k = 0
   18 - 6{{{(5/2)}}} - 6k = 0
   18 - 15 - 6k = 0
         3 - 6k = 0
            -6k = -3
              k = {{{(-3)/(-6)}}}
              k = {{{1/2}}}

Substituting in

        h² + k² = r²
        {{{(5/2)^2}}}{{{""+""}}}{{{(1/2)^2}}}{{{""=""}}}{{{r^2}}}
        {{{25/4}}}{{{""+""}}}{{{1/4}}}{{{""=""}}}{{{r^2}}}
                             {{{26/4}}}{{{""=""}}}{{{r^2}}}
                             {{{13/2}}}{{{""=""}}}{{{r^2}}}

So the equation

      (x - h)² + (y - k)² = r² 

becomes

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

It was harder but mainly because the center and radius were fractions.

Here is the graph.  Notice that the circle passes through the
given points and has center ({{{5/2}}},{{{1/2}}}) or (2.5,.5)

The radius is {{{sqrt(13/2)}}} &#8776; 2.55 

{{{drawing(400,400,-2,7,-4,5,grid(1), circle(5/2,1/2,sqrt(26)/2+.07),
circle(0,0,.2),circle(3,3,.2), circle(5,0,.2),
locate(0+.1,0+.5,"(0,0)"),
locate(3+.1,3+.5,"(3,3)"),
locate(5+.1,0+.5,"(5,0)"),
circle(5/2,1/2,.1), locate(5/2+.2,1/2+.5,C(5/2,1/2)),
circle(5/2,1/2,.08),circle(5/2,1/2,.1),circle(5/2,1/2,.05)


)}}}

Edwin</pre>