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. **

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