So I managed this question, but I don't the method I used was very effiecentWatch
Method 1: Note .
If you have P and A, that's enough to make it easy to find Q, (I'm not sure what you've worked out already).
Method 2: You should be able to write down a vector in the direction perpendicular to the first tangent (check that it would be going "towards" the circle and not away). If the circle has radius r, the point Q is a distance r from A in the direction of this vector.
Hence you can work out the coordinates of the point where the 2nd tangent touches the circle using basic vectors.
And finally substitute these coordinate values to find the value of k.
Finally you can then substitute the coordinate values in the equation of the tangent to work out the value of k.
Ok, I understand what you're saying, but can I just throw this in, if the gradient of both lines weren't both 2 would I still be able to do this question through geometry or would I have to sub in, and also is the same gradient the reason why I can assume that PA=AQ