Question 388149
<pre><font face = "batangche" color = "indigo"><b>
Let the center of the circle be O.


Draw in a radius to one of the points of tangency.  I'll draw OA.

{{{drawing(400,400,-8,11,-9.5,9.5, 

circle(0,0,7), locate(0,0,O),locate(10,0,P), locate(5,5.5,A),
line(0,0,10,0), line(10,0, -12,21.5), line(10,0, -12,-21.5),
green(line(0,0,4.9,4.9989999)), locate(4.8,-5,B), locate(2,3.2,7),
locate(4.5,0,10) 
 )}}}

OA is perpendicular to AP because a radius drawn to the point
of tangency is perpendicular to the tangent line.

Therefore triangle OAP is a right triangle. Therefore we can use 
the Pythagorean theorem:

{{{OA^2+AP^2=OP^2}}}
{{{7^2+AP^2=10^2}}}
{{{49+AP^2=100}}}
{{{AP^2=51}}}
{{{AP=sqrt(51)}}}

BP is also {{{sqrt(51)}}}

Edwin</pre>