<PRE><B>Question 983226</B>


The triangle &amp;nbsp;&lt;B&gt;OPX&lt;/B&gt;&amp;nbsp; is right-angled triangle because the radius drawn to the tangent point is perpendicular to the tangent. 

Therefore the area of the triangle  &amp;nbsp;&lt;B&gt;OPX&lt;/B&gt;&amp;nbsp; is half-product of its legs &amp;nbsp;&lt;B&gt;OP&lt;/B&gt;&amp;nbsp; and &amp;nbsp;&lt;B&gt;PX&lt;/B&gt;:


S = 12 = {{{1/2}}}.|&lt;B&gt;OP&lt;/B&gt;|.|&lt;B&gt;OP&lt;/B&gt;| = {{{1/2}}}*|&lt;B&gt;OP&lt;/B&gt;|*{{{6}}} = 3*|&lt;B&gt;OP&lt;/B&gt;|.


It gives for the radius of the circle &amp;nbsp;&amp;nbsp;r = |&lt;B&gt;OP&lt;/B&gt;| = {{{12/3}}} = 4.


Hence, &amp;nbsp;the area of the circle is &amp;nbsp;{{{pi*r^2}}} = {{{pi*4^2}}} = {{{16pi}}}.


&lt;B&gt;Answer&lt;/B&gt;. &amp;nbsp;The area of the circle is &amp;nbsp;{{{16pi}}}.


</PRE>