P is a point on a circle with the equation x^2 + y^2 = 45. P has x-coordinate 3 and is above the x axis. Work out the equation of the tangent to the circle at point P.

First insert x = 3 into the equation of the circle3^2 + y^2 = 459 + y^2 = 45 take 9 away from both sidesy^2 = 36 take the square root of both sidesy = 6 (not -6 as P is above the x-axis)Next find the gradient of the tangent. First find the gradient of the normal. Since the centre of the circle is at (0,0), the gradient (change in y/ change in x) is 6/3 = 2. This means that the gradient of the tangent is -1/2 (the negative reciprocal of the gradient of the normal).Finally, to find the equation of the tangent at P, use the formula y + y' = m(x - x').y + 6 = -1/2(x - 3) expand and simplify to the form y = mx + cy = -1/2x - 9/2 (or to remove fractions 2y + x = -9)