Question 1168231

given:

 tangent  line : {{{-x+y+4=0}}} or {{{y=x-4}}}
the center  ({{{1}}},{{{ 1}}})=({{{h}}},{{{ k}}})

since the distance between the point ({{{3}}}, {{{-1}}}), and the center  ({{{1}}}, {{{1}}}) is equal to the radius, we have

{{{r=sqrt((3-1)^2+(-1-1)^2)=sqrt(4+4)=sqrt(8)}}}


the equation of a circle is:

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

{{{(x-1)^2+(y-1)^2=(sqrt(8))^2}}}

{{{(x-1)^2+(y-1)^2=8}}}


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