Consider a right triangle, you can then use the fact that the tangent of the angle is equal to the opposite side over the adjacent side. So for a general point (a,b), we have thatOriginally Posted byConfuzzled?tan(x) = b/asox = arctan(b/a)where x is the angle you're looking for and "arctan" is the inverse tangent.

Well, find the intersection points! It's a system of two equations: solve the lineair one for either y or x (it's already solved for y) and substitute that expression in the equations of the circle. You then have a quadratic equation in 1 unknown which you can solve.Originally Posted byConfuzzled?

To find the perp bisector, you can set up the equation of a line through a point (the point P which is (A+B)/2, the middle of A and B where A and B are the two intersection points you just found) and with a given slope. You can find the slope because for two slopes c and d, we have thatcd = -1if they're perpendicular, and you're give the slope of the line of the chord.

It's a bit unclear to me, are they asking to show that the angle between AP and BP is 90°?Originally Posted byConfuzzled?