The circle C has centre (2,1) and radius 10. The point A(10,7) lies on the circle. Find the equation of the tangent to C at A and give it in the form 0 =ay + bx + c.
Whenever possible it is always a good idea to draw a graph to give a better understanding of what is going on in the question. As we have already been given a point on the tangent, namely A(10,7), it is a good idea to use the formula of a line y - y1 = m( x - x1) to work out the equation. Remember that here (x1,y1) is any point on the line a m is the gradient of the line. We know that one point on our tangent line is A(10,7). So our first step is to substitute x1 = 10 and y1 = 7, giving us y - 7 = m (x - 10). Now all that is left to do is find the gradient of the tangent, m. We know that the radius of the circle is perpendicular to the tangent. Also we know that to find the gradient of a line which is perpendicular to another line we simply take the negative reciprocal of the gradient of the original line. So we must find the gradient of the radius. To do this we simply take the difference in y coordinates and divide them by the difference in x coordinates. ie (7-1)/(10-2) = 6/8. This is easy to see from our diagram. Now, as mentioned before we must take the negative reciprocal of 6/8 and this will give us the gradient of the tangent, thus m = -8/6.Now we go back to our original formula subbing in our value for m. Giving us y - 7 = -8/6 (x - 10). This is not quite the final answer however, as we must write it in the form given in the question. To do this we simply rearrange. We do this by multiplying both sides by 6, expanding the brackets and bringing everything onto one side. Giving us our final answer of 0 = 6y - 8x -122.
