Question 968506
Center at the point ({{{-4}}},{{{-3}}}); containing the point ({{{-3}}},{{{3}}})

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

if the center is at the point ({{{h}}},{{{k}}})= ({{{-4}}},{{{-3}}}),  we 

{{{(x-(-4))^2+(y-(-3))^2=r^2}}}

{{{(x+4)^2+(y+3)^2=r^2}}}..........eq.1

now use given point ({{{x}}},{{{y}}})=({{{-3}}},{{{3}}})

{{{(-3+4)^2+(3+3)^2=r^2}}}.........solve for {{{r^2}}}

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

{{{1+36=r^2}}}

{{{r^2=37}}}

{{{r=sqrt(37)}}}

then, your equation is: {{{(x+4)^2+(y+3)^2=37}}}


{{{drawing( 600,600, -15, 15, -15, 15,
circle(-4,-3,.12),circle(-3,3,.12),circle(-4,-3,sqrt(37)),
locate(-4,-3,p(-4,-3)),locate(-3,3,p(-3,3)),
 graph( 600,600, -15, 15, -15, 15, 0)) }}}