**Problem**: Given a circle that passes through the points $(3,4)$ and $(6,8)$, find the length of a tangent to the circle from the origin.

**Solution**: Consider Power of a Point, which states that

$a^2 = b(b+c) = d(d+e)$, all of which are equal to the

*power*of the point at which they intersect.

Now consider the tangent from the origin in the problem (call this length $x$) and notice that the secant from the origin through $(3,4)$ passes through $(6,8)$ as well.

Then by Power of a Point applied to the origin, we have $x^2 = (\sqrt{3^2+4^2})(\sqrt{6^2+8^2}) = (5)(10) = 50 \Rightarrow x = \sqrt{50}$. QED.

Note: On the actual contest (ARML) a third point was provided to define the circle. We notice, however, that this third point is not necessary and was thus omitted in this rewording of the problem.

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Practice Problem: Given two circles, find all points $P$ such that the power of point $P$ with respect to both circles is equal.

Wow....great math blog! I am going to check this every day, so update (or face the wrath of me :-))!

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