# 1959 AHSME Problems/Problem 32

## Problem

The length $l$ of a tangent, drawn from a point $A$ to a circle, is $\frac43$ of the radius $r$. The (shortest) distance from A to the circle is: $\textbf{(A)}\ \frac{1}{2}r \qquad\textbf{(B)}\ r\qquad\textbf{(C)}\ \frac{1}{2}l\qquad\textbf{(D)}\ \frac23l \qquad\textbf{(E)}\ \text{a value between r and l.}$

## Solution

$[asy] import geometry; point O=(0,0); point A=(5,0); point B,T; circle c=circle(O,3); markscalefactor=0.05; // Circle, segment OA draw(c); dot(O); label("O",O,NW); dot(A); label("A",A,NE); draw(O--A); // Segments OT, OA line[] t1=tangents(c,A); pair[] t=intersectionpoints(t1[0], c); T=t[0]; dot(t[0]); label("T",T,SE); draw(A--T--O); draw(rightanglemark(A,T,O)); // Point B pair[] b=intersectionpoints((O--A),c); B=b[0]; dot("B",B,NE); // Length labels label("r",midpoint(O--T),SW); label("r",midpoint(O--B),N); label("\frac{4}{3}r",midpoint(A--T),SE); [/asy]$

Let the circle have center $O$, let the point of tangency be point $T$, and let $B$ be the intersection of $\overline{OA}$ with the circle, as in the diagram. By the definitions of a circle and a tangent to a circle, we know that $OB=OT=r$ and $\overline{OT} \perp \overline{TA}$. By the Pythagorean Theorem, $OA=\sqrt{\frac{25r^2}{9}}=\frac{5}{3}r$. Because the shortest segment from an external point to a circle lies on the line connecting that point to the center of the circle, our desired distance is $AB$. Because $OA=\frac{5}{3}r$ and $OB=r$, $AB=\frac{5}{3}r-r=\frac{2}{3}r=\frac{(4/3)r}{2}=\boxed{\textbf{(C) }\frac{1}{2}l}$.