Answer
See below
Work Step by Step
Given $x^2+y^2-14x+4y-11=0$
We can see that $a=1\\b=0\\c=1$
We will find the discriminant of the given equation $=b^2-4ac\\=0^2-4(1)(1)\\=-4$
Since $-4\lt0$ and $a=c=1$, the conic is a circle.
To graph the circle, first complete the square:
$x^2+y^2-14x+4y-11=0\\x^2-14x+y^2+4y=11\\(x^2-14x+49)-49+(y^2+4y+4)-4=11\\(x-7)^2+(y+2)^2=64$
From the equation, you can see that the center is $(7,-2)$ and the radius is $r=8$.
