Question 985218


same as Question 985229

Find the center and the radius of the circle that passes through the points 

({{{4}}},{{{4}}}) , ({{{1}}},{{{3}}}) and ({{{8}}},{{{-4}}}).


equation of  the circle is:

{{{(x-h)^2+(y-k)^2=r^2}}} where {{{h}}} and {{{k}}} are {{{x}}} and {{{y}}}  coordinates of the center, and {{{r}}} is radius

so use given points to set up system of three unknown 
({{{4}}},{{{4}}})
{{{(4-h)^2+(4-k)^2=r^2}}}....simplify

{{{16-8h+h^2+16-8k+k^2=r^2}}}.............eq.1

 ({{{1}}},{{{3}}})

{{{(1-h)^2+(3-k)^2=r^2}}}

{{{1-2h+h^2+9-6k+k^2=r^2}}}.............eq.2


({{{8}}},{{{-4}}})

{{{(8-h)^2+(-4-k)^2=r^2}}}....simplify

{{{64-16h+h^2+16+8k+k^2=r^2}}}.............eq.3

start with
{{{16-8h+h^2+16-8k+k^2=r^2}}}.............eq.1
{{{64-16h+h^2+16+8k+k^2=r^2}}}.............eq.3
--------------------------------------------------------------------subtract

{{{16-8h+h^2+16-8k+k^2-(64-16h+h^2+16+8k+k^2)=r^2-r^2}}}
{{{cross(16)-8h+cross(h^2)+16-8k+cross(k^2)-64+16h-cross(h^2)-cross(16)-8k-cross(k^2)=r^2-r^2}}}

{{{-8h+16-8k-64+16h-8k=0}}}

{{{8h-16k=64-16}}}.......both sides divide by 8

{{{h-2k=8-2}}}

{{{h-2k=6}}}...............eg.1a


go with 
{{{16-8h+h^2+16-8k+k^2=r^2}}}.............eq.1
{{{1-2h+h^2+9-6k+k^2=r^2}}}.............eq.2
---------------------------------------------------------subtract

{{{16-8h+h^2+16-8k+k^2-(1-2h+h^2+9-6k+k^2)=r^2-r^2}}}

{{{16-8h+cross(h^2)+16-8k+cross(k^2)-1+2h-cross(h^2)-9+6k-cross(k^2)=0}}}

{{{22-8h-8k+2h+6k=0}}}


{{{-6h-2k=-22}}}...............eg.2a

now go with
{{{h-2k=6}}}...............eg.1a
{{{-6h-2k=-22}}}...............eg.2a
-----------------------------------------------subtract 

{{{h-2k-(-6h-2k)=6-(-22)}}}
{{{h-cross(2k)+6h+cross(2k)=6+22}}}
{{{h+6h=28}}}
{{{7h=28}}}
{{{h=28/7}}}
{{{highlight(h=4)}}}

go to 
{{{h-2k=6}}}...............eg.1a...plug in {{{4}}} for {{{h}}}

{{{4-2k=6}}}
{{{4-6=2k}}}
{{{-2=2k}}}
{{{k= -2/2}}}
{{{highlight(k= -1)}}}

go to

{{{16-8h+h^2+16-8k+k^2=r^2}}}.............eq.1 ..plug in {{{4}}} for {{{h}}} and {{{-1}}} for {{{k}}} and find {{{r}}}

{{{16-8(4)+4^2+16-8(-1)+(-1)^2=r^2}}}

{{{cross(16)-cross(32)+cross(16)+16+8+1=r^2}}}

{{{16+8+1=r^2}}}

{{{25=r^2}}}

{{{highlight(r=5)}}}

so, your equation is: {{{(x-4)^2+(y+1)^2=25}}} 


check:


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(4,-1,.12),locate(4,-1,C(4,-1)),
circle(4,-1,5),
circle(8,-4,.12),locate(8,-4,p(8,-4)),
circle(4,4,.12),locate(4,4,p(4,4)),
circle(1,3,.12),locate(1,3,p(1,3)),
 graph( 600, 600, -10, 10, -10, 10, 0)) }}}