Find the equation of the tangent to the circle (x-3)^2 + (y-4)^2 = 13 that passes through the point (1,7)

Start with a sketch. We can see that the radius from the point (1,7) to the centre of the circle (3,4) is perpendicular to the tangent. The gradient of the radius is (4-7)/(3-1) = -3/2. We know that two perpendicular gradients multiply to make -1, so the gradient, m, of the tangent is 2/3.
The equation of the tangent is now y=2/3x + c . To find c, all we have to do is plug in a coordiante for x and y - we know (1,7) lies on the tangent so we will use this. Therefore c=19/3. The equation of our tangent is therefore y=2/3 x + 19/3 !