Question 1076714
 Find an equation of the circle that passes through the points ({{{2}}},{{{3}}}), ({{{4}}},{{{5}}}), and ({{{0}}},{{{-3}}}). 

The standard form equation of a circle is {{{ (x-h)^2+(y-k)^2=r^2}}} where

{{{h}}} and {{{k}}} are the {{{x}}} and {{{y}}} coordinates of the center of the circle 

first find  coordinates of the center  ({{{h}}},{{{k}}}) using given points

The first step is to set up these 3 equations by plugging the x- and y-coordinates of the points 
({{{2}}},{{{3}}}), ({{{4}}},{{{5}}}), and ({{{0}}},{{{-3}}})  into the circle formula:

{{{(2 - h)^2 + (3 - k)^2 = r^2}}}.................eq1
{{{(4 - h)^2 + (5 - k)^2 = r^2}}}..................eq2
{{{(0 - h)^2 + (-3 - k)^2 = r^2}}}.................eq3
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{{{h^2 - 4 h + k^2 - 6 k + 13 = r^2}}}.................eq1
{{{h^2 - 8 h + k^2 - 10 k + 41= r^2}}}..................eq2
{{{h^2 + k^2 + 6 k + 9 = r^2}}}...............eq3

--------------------------------------------


{{{h^2 - 4 h + k^2 - 6 k + 13 = r^2}}}.................eq1
{{{h^2 - 8 h + k^2 - 10 k + 41= r^2}}}..................eq2...............subtract eq2 from eq1
---------------------------------------------------------
{{{h^2 - 4 h + k^2 - 6 k + 13-(h^2 - 8 h + k^2 - 10 k + 41)= r^2-r^2}}}
{{{h^2 - 4 h + k^2 - 6 k + 13-h^2 +8 h- k^2 +10 k - 41= 0}}}
{{{4 h + 4 k - 28= 0}}}............simplify
{{{ h +  k - 7= 0}}}.............eq1a


h^2 - 4h + k^2 - 6 k + 13 = r^2.................eq1
h^2 + k^2 + 6k + 9 = r^2...............eq3...........subtract eq3 from eq1
------------------------------------------
{{{h^2 - 4h + k^2 - 6k + 13  -(h^2 + k^2 + 6 k + 9 )= r^2-r^2}}}
{{{h^2 - 4h + k^2 - 6k + 13  -h^2 - k^2 - 6 k - 9 =0}}}
{{{-4 h - 12k + 4 =0}}}............divide by -4
{{{ h +3k - 1 =0}}}..............................eq2a

use 
 {{{h +  k - 7= 0}}}...............eq1a
 {{{ h +3 k - 1 =0}}}..............................eq2a
-------------------------------------------------------------subtract eq1a from eq2a

{{{ h +3 k - 1 -(h +  k - 7)=0}}}

 {{{h +3 k - 1 -h -  k + 7=0}}}

{{{2 k  + 6=0}}}

{{{2 k  =-6}}}

{{{ highlight(k  = -3)}}}

now find {{{h}}}

 {{{h +  k - 7= 0}}}...............eq1a

 {{{h -3- 7= 0}}}

{{{h-10=0}}}

{{{h=10}}}


so, center is at  ({{{10}}},{{{-3}}})

use one point, h and k to find r

{{{(2 - 10)^2 + (3 - (-3))^2 = r^2}}}.................eq1

{{{( - 8)^2 + (3 +3)^2 = r^2}}}

{{{64 + 36 = r^2}}}

 {{{r^2=100}}}

{{{r=10}}}

and, your equation is:

 {{{(x-10)^2+(y+3)^2=10^2}}}

(2,3), (4,5), and (0,-3).
{{{drawing( 600, 600, -20, 20, -20, 20,  
circle(10,-3,.13),locate(10,-2.8,C(10,-3)),
circle(2,3,.13),locate(2,2.8,p(2,3)),
circle(4,5,.13),locate(4,4.8,p(4,5)),
circle(0,-3,.13),locate(0,-3,p(0,-3)),
circle(10,-3,10),
graph( 600, 600, -20, 20, -20, 20, 0)) }}}