Differentiation gives us the gradient dy/dx: 2x+2ydy/dx=0, so dy/dx=-x/y.
Plug in the point: dy/dx=1/5, which is the slope at that point. However the point doesn't lie on the circle, because (-1)^2+5^2=26, so we must assume that the equation of the circle should be x^2+y^2=26.
On this assumption, then, we know the slope is 1/5 and the equation of the tangent is y=x/5+a where a is to be found. Plug in (-1,5): 5=-1/5+a, so a=26/5, and y=x/5+26/5 or 5y=x+26.