sHow that x^2-6x-39=0

Question

AT is a tangent to a circle, centre O. OT = X cm, AT = (X+5) cm and OA = ( X+8) cm. Point A is outside the circle at the end of the tangent. sHow that x^2-6x-39=0

Answer 1

This is simply about a right triangle (rt. angle at T) because the radius to a point of tangency is ⊥ the tangent.

 

Then (x+8)² = x² + (x+5)² and this in fact reduces to x² - 6x -39 = 0.