Question 1087291
equation of the circle is:

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

if a circle has center ({{{-1}}},{{{2}}}), means  {{{h=-1}}} and {{{k=2}}}

so far we have:

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

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

and if passes through ({{{-5}}},{{{6}}}), use it to find {{{r}}}

{{{(-5+1)^2+(6-2)^2=r^2}}}

{{{(-4)^2+(4)^2=r^2}}}

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

{{{r^2=32}}}

so, your equation is:

{{{(x+1)^2+(y-2)^2=32}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-1,2,.12),circle(-5,6,.12),
locate(-1,2,C(-1,2)),locate(-5,6,p(-5,6)),
 graph( 600, 600, -10, 10, -10, 10,-sqrt(32-(x+1)^2)+2, sqrt(32-(x+1)^2)+2))) }}}