Answer
See below
Work Step by Step
Given: $x^2+y^2+4x-14y+17=0$
Because $A=1,B=0$, and $C=1$, the discriminant is $B^2-4AC=-4 \lt 0$, so the conic is a circle. Complete the square to write the equation in standard form.
$x^2+y^2+4x-14y+17=0\\(x^2+4x+4)-4+(y^2-14y+49)-49+17=0\\(x+2)^2+(y-7)^2=36$
From the equation, we can see the center is at $(h,k)=(-2,7)$ and the radius is $r=6$.
The graph is shown below.
