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Circle $O$ is tangent to $\overline{AB}$ at $A$, and $\angle ABD = 90^\circ$. If $AB = 12$ and $CD = 18$, find the radius of the circle.

 Mar 18, 2023
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Let $r$ be the radius of the circle, $O$ be the center of the circle, and $E$ be the foot of the altitude from $O$ to $\overline{BD}$, as shown below.

 

Since $O$ is the center of the circle and $A$ is a point on the circle, we have $OA = r$. By the Pythagorean Theorem applied to right triangle $ABD$, we have $BD^2 = AB^2 + AD^2 = 12^2 + r^2$. By the Pythagorean Theorem applied to right triangle $OBD$, we have $OD^2 = OB^2 + BD^2 = (r + 12)^2 + BD^2$. But $OD = OE + ED = r + DE$, so we have \begin{align*} (r + DE)^2 &= (r + 12)^2 + BD^2 \ r^2 + 2rDE + DE^2 &= r^2 + 24r + 144 + BD^2 \ 2rDE &= 24r + 144 - BD^2 - DE^2 \ DE &= \frac{12r + 72 - BD^2/2}{r}. \end{align*} By the Pythagorean Theorem applied to right triangle $OBE$, we have $OE^2 = OB^2 - BE^2 = r^2 - (\frac{BD}{2})^2$. But $OE = DE - DO = DE - (r + 12)$, so we have \begin{align*} (DE - (r + 12))^2 &= r^2 - \frac{BD^2}{4} \ DE^2 - 2(r + 12)DE + (r + 12)^2 &= r^2 - \frac{BD^2}{4} \ DE^2 - 2(r + 12)DE &= r^2 + \frac{BD^2}{4} - (r + 12)^2 \ DE &= \frac{r^2 + \frac{BD^2}{4} - (r + 12)^2}{2(r + 12)} \ &= \frac{r^2 - 24r + 72}{2(r + 12)} + 9. \end{align*} Setting the two expressions for $DE$ equal, we have [ \frac{12r + 72 - BD^2/2}{r} = \frac{r^2 - 24r + 72}{2(r + 12)} + 9. ]

Solving for $r$, we get $r = 14$.  The radius of the circle is 14.

 Mar 18, 2023

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