<PRE><B>Question 1209461</B>
&lt;br&gt;
The graph of x^2+y^2=4 is a circle with center at the origin and radius 2.&lt;br&gt;
Let O be the origin; let A and B be the y- and x-intercepts of the line y=mx+c; and let C be the point of tangency with the circle.&lt;br&gt;
AOB is a right triangle, and OC is the altitude to the hypotenuse.&lt;br&gt;
The y- and x-intercepts of the given line are&lt;br&gt;
A(0,c)
B(c/m,0)&lt;br&gt;
That gives us the lengths of the legs of triangle AOB as&lt;br&gt;
OA = c
OB = c/m&lt;br&gt;
In right triangle ACO, OA = c and OC = 2, so AC = sqrt(c^2-4).&lt;br&gt;
Triangles AOB and ACO are similar.  Set up and solve a proportion using the lengths of the legs of those two triangles.&lt;br&gt;
{{{2/sqrt(c^2-4)=(c/m)/c}}}&lt;br&gt;
{{{2/sqrt(c^2-4)=1/m}}}
{{{4/(c^2-4)=1/m^2}}}
{{{c^2-4=m^2}}}&lt;br&gt;
QED....&lt;br&gt;
</PRE>