C is a circle with center O. AB is a chord not passing through O. M is the midpoint of AB. C' is the circle with diameter OM. T is a point on C'. The tangent to C' at T meets C at P. Show that PA² + PB² = 4 PT².
Set up a Cartesian coordinate system with the origin at O. We choose
our axes such that the x-axis is parallel to the chord AB, and our
units such that circle C has radius 1.
Our coordinates are then:
O - (0,0)
A - (-cos x, sin x)
B - (cos x, sin x)
M - (0, sin x)
P - (cos y, sin y)
X - (0, ½ sin x)
where X is the centre of the smaller circle
Then
PA² = (cos y + cos x)² + (sin y - sin x)²
PB² = (cos y - cos x)² + (sin y - sin x)²
PT² = PX² - XT² = (cos y)² + (sin y - ½ sin x)² - OX² (since XT and OX are the radii of the smaller circle)
= (cos y)² + (sin y - ½ sin x)² - (½ sin x)²
Just expand and it's easy to see how the equality holds...
Ugly solution
