Point U is outside of a circle. Points X and Y are on the circle and are colinear with point P (i.e., the three points are on the same line)

Let O be the center of the circle, and let r be the radius of the circle.

Since line UV is tangent to the circle at point V, we know that UV=r.

We also know that UX=r+x and UY=r+y, where x and y are the distances from points U and V to the line XY, respectively.

Since XY is a straight line, we know that x+y=r.

Substituting r+x for UX and r+y for UY in the equation UV2=UX∗UY, we get:

(r)2=(r+x)(r+y)

r2=r2+rx+ry+xy

0=rx+ry+xy

Since x+y=r, we can substitute r for x+y in the equation 0=rx+ry+xy to get:

0=r(r−y)+xy

0=r2−yr+xy

We can factor the equation 0=r2−yr+xy to get:

0=(r−y)(r+y)

Since x+y=r, we know that y=r−x. Substituting r−x for y in the equation 0=(r−y)(r+y), we get:

0=(r−(r−x))(r+(r−x))

0=(2r−x)(2r−x)

0=(2r−x)2

Since the square of a real number is always non-negative, we know that (2r−x)2=0. Therefore, 2r−x=0.

Solving for x, we get:

x=2r

Substituting 2r for x in the equation UV2=UX∗UY, we get:

UV2=(r+2r)(r+2r)

UV2=4r2

Therefore, UV2=UX∗UY.