For a function f(x), the slope of the tangent at any point x = a is given by f'(a). If the tangent is a vertical line the slope is of the form C/0.
The circle has the equation x^2 + y^2 - 8 = 0
using implicit differentiation
2x + 2y*y' = 0
=> y' = -x/y
As the slope of the tangent should have the denominator as 0, y = 0
substituting y = 0 in x^2 + y^2 - 8 = 0
=> x^2 - 8 = 0
=> x^2 = 8
=> x = sqrt 8 and x = -sqrt 8
The tangents drawn to the circle x^2 + y^2 - 8 = 0 at the points (2*sqrt 2, 0) and (-2*sqrt 2, 0) are vertical lines.