Tangent and secant from an external point

Text

From a point $P$ external to a circle of centre $O$ and radius $2$ you draw a tangent, touching the circle at $A$. From $P$ you also draw a secant meeting the circle first at $B$, then at $C$, so that the diameter $AD$ is met by $PC$ at a point $E$ which is the midpoint of the radius $OD$. You know that $\overline{AP}=4$. Determine $PB$, $BE$, $EC$. (figure in the original test: tangent $PA$ perpendicular to the diameter $AD$; the secant $PBC$ passes through $E$, the midpoint of $OD$.)

Solution

Set coordinates $O(0;0)$, $A(0;2)$ (tangency point at the top), $D(0;-2)$. The tangent at $A$ is horizontal and $\overline{AP}=4$, so $P(4;2)$. $E$ is the midpoint of $OD$: $E(0;-1)$. The secant is line $PE$: from $P(4;2)$ toward $E(0;-1)$, direction $(-4;-3)$ of magnitude $5$; parametrize $(x;y)=(4-4t;\,2-3t)$. Intersection with the circle $x^2+y^2=4$: $(4-4t)^2+(2-3t)^2=4\ \Longrightarrow\ 25t^2-44t+16=0\ \Longrightarrow\ t=\frac{22\pm2\sqrt{21}}{25}.$ Distances from $P$ equal $5t$; $B$ is the first intersection ($smaller$ $t$), $C$ the second: $PB=\frac{22-2\sqrt{21}}{5}\approx2{,}57,\qquad PC=\frac{22+2\sqrt{21}}{5}\approx6{,}23.$ Check with the power of $P$: $PB\cdot PC=\overline{AP}^2=16$ ✓. Since $E$ corresponds to $t=1$, $\;PE=5$, so $BE=PE-PB=\frac{3+2\sqrt{21}}{5}\approx2{,}43,\qquad EC=PC-PE=\frac{2\sqrt{21}-3}{5}\approx1{,}23.$ $\boxed{PB=\tfrac{22-2\sqrt{21}}{5}\approx2{,}57,\quad BE=\tfrac{3+2\sqrt{21}}{5}\approx2{,}43,\quad EC=\tfrac{2\sqrt{21}-3}{5}\approx1{,}23.}$