4x^2+y^2-32x+60=0
First complete the square:
$\displaystyle \Rightarrow 4(x^2 - 8x) + y^2 + 60 = 0 \Rightarrow 4([x - 4]^2 - 16) + y^2 + 60 = 0 \Rightarrow 4(x - 4)^2 + y^2 = 4$.
Now re-arrange into standard form to see coordinates of centre, lentgth of axes etc.
Use the original form to get x- and y-intercepts.