Question 650185
{{{(x - h)^ 2 + (y - k)^ 2 = r ^2}}}

you have  {{{(x+1)^2 + (y-4)^2=9}}}

so, {{{h=-1}}}, {{{k=4}}}, and {{{r^2=9}}}...->{{{r=3}}}

center at {{{C(h , k) = C(-1 , 4) }}} and radius    {{{r = 3}}}

To find the {{{x-intercepts}}}, set {{{y = 0}}} in the above equation and solve 
for {{{x}}}. 

{{{(x+1)^2 + (0-4)^2=9}}}

{{{(x+1)^2 + (-4)^2=9}}}

{{{(x+1)^2 +16=9}}}

{{{(x+1)^2 =9-16}}}

{{{(x+1)^2 =-7}}}

{{{sqrt((x+1)^2 )=sqrt(-7)}}}

{{{x+1 =sqrt(-7)}}}

{{{x =sqrt(-7)-1}}}

as you can see,  there is {{{none}}} of {{{x- intercepts }}} because there is no real solution

To find the {{{y- intercepts}}}, set {{{x = 0}}} in the above equation and solve 
for {{{y}}}. 

{{{(0+1)^2 + (y-4)^2=9}}}

{{{(1)^2 + (y-4)^2=9}}}

{{{1 + (y-4)^2=9}}}

{{{(y-4)^2=9-1}}}

{{{(y-4)^2=8}}}

{{{sqrt((y-4)^2)=sqrt(8)}}}

{{{y-4=sqrt(8)}}}.........{{{sqrt(8)=2.8}}} and {{{sqrt(8)=-2.8}}}

{{{y=2.8+4}}}

{{{y=6.8}}}

and
{{{y=-2.8+4}}}

{{{y=1.2}}}


{{{y- intercepts }}} are {{{6.8}}} and {{{1.2}}}


{{{drawing(300,300,-5,10,-5,10,grid(1),circle(-1,4,3))}}}