C is a circle with equation x^2 + y^2 = 16. The point P = (3,√7) is on C. Find the equation of the tangent to C at the point P.

First, we note the centre of the circle is at the origin (0,0). Then we calculate the gradient of the line OP, the line connecting the origin to P. The gradient = change in y / change in x = (√7 - 0) / (3 - 0) = √7 / 3. We know that m1 x m2 = -1 when m1 and m2 are the gradients of perpendicular lines. Hence the gradient of the tangent to C at P is -1 / (√7 / 3) = -3 / √7.Writing the equation of the tangent in the form y = mx + c we know y = (-3 / √7)x + c. To find the value of c we use the values of x and y given by P as we know P is on this line. This gives us √7 = (-3 / √7)(3) + c and rearranging to make c the subject gives √7 + 9/√7 = c. Making √7 the common denominator we can see that c = 16/√7 and hence the equation of the tangent is y = (-3 / √7)x + 16/√7.