Question 1031604
<pre>
The standard equation of a circle is

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

Substituting the three points we have this system:

{{{system((-3-h)^2+(6-k)^2=r^2,(-5-h)^2+(2-k)^2=r^2,(3-h)^2+(-6-k)^2=r^2)}}}

Simplifying and rearranging the terms:

{{{system(h^2+6h+k^2-12k+45=r^2,h^2+10h+k^2-4k+29=r^2,h^2-6h+k^2+12k+45=r^2)}}}

Subtracting the third equation from the first:

{{{12h-24k=0}}}
{{{h-2k=0}}}
{{{h=2k}}}

Subtracting the second equation from the first:

{{{-4h-8k+16=0}}}
{{{h+2k-4=0}}}

Substituting 2k for h

{{{2k+2k-4=0}}}
{{{4k-4=0}}}
{{{4k=4}}}
{{{h=1}}}

So the center = (h,k) = (2,1)

Substituting h=2 and k=1 in the first equation

{{{h^2+6h+k^2-12k+45=r^2}}}
{{{2^2+6(2)+1^2-12(1)+45=r^2}}}
{{{4+12+1-12+45=r^2}}}
{{{50=r^2}}}
{{{sqrt(50)=r}}}
{{{5sqrt(2)=r}}}

The equation

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

{{{(x-2)^2+(y-1)^2=50}}}

Or if you multiply that out and get the general form:

{{{x^2+y^2-4x-2y+45=0}}}

{{{drawing(400,400, -10,12,-9,13,
graph(400,400, -10,12,-9,13),
circle(-3,6,0.15),circle(-3,6,0.13),circle(-3,6,0.11),circle(-3,6,0.09),circle(-3,6,0.07),circle(-3,6,0.05),circle(-3,6,0.03),circle(-3,6,0.01),
locate(-3,6,"(-3,6)"),locate(-5,2,"(-5,2)"),locate(3,-6,"(3,-6)"),
locate(2,1,"(2,1)"),
 
circle(-5,2,0.15),circle(-5,2,0.13),circle(-5,2,0.11),circle(-5,2,0.09),circle(-5,2,0.07),circle(-5,2,0.05),circle(-5,2,0.03),circle(-5,2,0.01),

circle(3,-6,0.15),circle(3,-6,0.13),circle(3,-6,0.11),circle(3,-6,0.09),circle(3,-6,0.07),circle(3,-6,0.05),circle(3,-6,0.03),circle(3,-6,0.01),

circle(2,1,0.15),circle(2,1,0.13),circle(2,1,0.11),circle(2,1,0.09),circle(2,1,0.07),circle(2,1,0.05),circle(2,1,0.03),circle(2,1,0.01),

circle(2,1,sqrt(50)) )}}}
Edwin</pre>