As the tangent meets the circle at P, when x =1, we can assume that the y value will be the same for this value of x. therefore we substitute x=1 into the equation of the cirlce. Rearranging the equation of the circle gives x = 3. Therefore the line that intersects P parallel to the tangent has the gradient 3. Therefore as it is perpendicular to the tangent, the tangent has the gradient -1/3. this gives the equation of the tangent as y = -(1/3)x + c. We therefore substitute in the two known values of y and x (coordinate P) to find the value of c (the constant). At point Q, y = 0. As we now have the equation y = -(1/3)x + 10/3 we can substitute in the value of y and rearrange for x to find the coordinates for q. This qives q as (10,0).