P is a point on the circle with equation x^2 + y^2 = 80. P has x-coordinate 4 and is below the x-axis.Work out the equation of the tangent to the circle at P.

Point P has x-coordinate 4, to work out the y-coordinate you should substitute x=4 into the equation of the circle. This gives you 16 + y^2 = 80. this can be rearranged to y^2 = 80 - 16 = 64. Therefore y = +8 or -8. The question tells us that P is below the x-axis so we know that the y-coordinate of P is -8. this is the only known point on the tangent so it is not yet enough to workout the equation of the tangent. However you we know that the radius of the circle through point P will be perpendicular to the tangent so if we workout the gradient of the radius through point P we will be able to workout the gradient of the tangent. The gradient of the radius is -8/4 which is equal to -2. Therefore the gradient of the tangent is 0.5. the general equation of the tanget is y = .5x + c, where c is a constant. to work out c you can sub in y = -8 and x = 4 as you know that point lies on the tangent. When you do this you get -8 = 2 + C this can be rearranged to give C = -8-2 = -10. Now you know c the equation the tangent can be given.Answer: y = .5x - 10