Question 294471
<pre><font size = 4 color = "indigo"><b>
{{{drawing(400,400,-13,13,-13,13,
locate(12.5,0,x), locate(0.3,12.8,y),
red(line(-13,0,13,0)), red(line(0,-13,0,13)),
circle(0,0,5), line(0,25/4,25/3,0),
green(line(0,0,3,4)), green(line(3,0,3,4)),
locate(0,0,O), locate(3,0,D), locate(25/3,0,A),
locate(3,4+1.5,T(x[1],y[1])), locate(0,25/4+1,B(0,b)),
green(locate(1.2,3,r)),locate(1.3,.3,x[1]),
locate(3,2.7,y[1])

  )}}}

AB is tangent to circle O, whose equation is {{{x^2+y^2=r^2}}}.
OT is a radius (of length r) drawn to T(x<sub>1</sub>,y<sub>1</sub>),
the point of tangency. Therefore OT is perpendicular to AB.
OB has length b, the y-coordinate of the y-intercept of AB and 
B is the point (0,b).  

All triangles in the figure above are right triangles, and
all of them are similar!  This is easy to see if you
realize that the pair of acute angles in each of them are
equal in measure.  In particular since triangles OTB and TDO
are similar,  

{{{(OB)/(OT)=(OT)/(TD)}}}

Observing their lengths:

{{{b/r=r/y[1]}}}

{{{b=r^2/y[1]}}}

Edwin</pre>