Points $A$, $B$, and $C$ are on a circle such that $AB = 8$, $BC = 15$, and $AC = 12$. Find the radius of the circle.
By the Law of Cosines on triangle $ABC$, we have
\begin{align*}
\cos \angle ABC &= \frac{8^2+15^2-12^2}{2\cdot 8 \cdot 15} = \frac{97}{120}\\
\cos \angle ABC &= \cos(\angle ABC + \angle CAB) \\
\cos \angle ABC &= \cos(\angle BAC) \cos(\angle ABC) - \sin(\angle BAC) \sin(\angle ABC) \\
2 \cos^2 \angle ABC - 1 &= (2 \cos(\angle ABC)^2 - 1) \cos(\angle BAC) - 2 \sin(\angle ABC) \sin(\angle BAC) \cos(\angle ABC) \\
153 \cdot \frac{97}{120} - 1 &= \left(\frac{97}{120}\right)^2 \cos(\angle BAC) - 2 \frac{120}{\sqrt{1441}} \cdot \frac{97}{120} \cdot \frac{77}{120}\\
\frac{11681}{80} &= \left(\frac{97}{120}\right)^2 \cos(\angle BAC) - \frac{5099}{96}\\
\frac{11681}{80} + \frac{5099}{96} &= \left(\frac{97}{120}\right)^2 \cos(\angle BAC)\\
\frac{77909}{240} &= \left(\frac{97}{120}\right)^2 \cos(\angle BAC)\\
160^2 &= 120^2 \cos^2(\angle BAC)\\
160^2 &= 120^2 - 120^2 \sin^2(\angle BAC)\\
s &= \sqrt{120^2 - 120^2 \sin^2(\angle BAC)}
\end{align*}Since $\angle BAC$ is acute,
\[\sin \angle BAC = \sqrt{1 - \cos^2 \angle BAC} = \sqrt{1 - \left( \frac{160}{120} \right)^2} = \frac{3}{\sqrt{7}}.\]Thus,
\[s = \sqrt{120^2 - 120^2 \cdot \frac{9}{7}} = \sqrt{120^2 \cdot \frac{16}{7}} = \boxed{\frac{240}{\sqrt{7}}}.\]
To find the radius of the circle, we can use the formula for the radius of a circle inscribed in a triangle.
The formula is: $r = \frac{A}{s}$, where $r$ is the radius, $A$ is the area of the triangle, and $s$ is the semiperimeter of the triangle.
In this case, we can use Heron's formula to find the area of the triangle.
Heron's formula states that the area $A$ of a triangle with side lengths $a$, $b$, and $c$ is given by:
\[A = \sqrt{s(s-a)(s-b)(s-c)}\]
where $s$ is the semiperimeter of the triangle, given by:
\[s = \frac{a+b+c}{2}\]
In this case, the side lengths of the triangle are $a = 8$, $b = 15$, and $c = 12$, so we can find the semiperimeter as follows:
\[s = \frac{a+b+c}{2} = \frac{8+15+12}{2} = \frac{35}{2}\]
Now, we can use Heron's formula to find the area of the triangle:
\[A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{35}{2}\left(\frac{35}{2}-8\right)\left(\frac{35}{2}-15\right)\left(\frac{35}{2}-12\right)}\]
\[= \sqrt{\frac{35}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{23}{2}}=\sqrt{\frac{35\cdot7\cdot5\cdot23}{2\cdot2\cdot2\cdot2}}\]
\[= \sqrt{\frac{35\cdot7\cdot5\cdot23}{2^4}}=\sqrt{\frac{35\cdot7\cdot5\cdot23}{16}}.\]
Therefore, the area of the triangle is $\sqrt{\frac{35\cdot7\cdot5\cdot23}{16}}$.
Now we can use the formula for the radius of a circle inscribed in a triangle to find the radius of the circle:
\[r = \frac{A}{s}=\frac{\sqrt{\frac{35\cdot7\cdot5\cdot23}{16}}}{\frac{35}{2}}.\]
Finally, simplifying the expression above gives the radius of the circle.
To find the radius of the circle, we can use the properties of a circle. In this case, we will use the fact that the lengths of two chords that intersect inside a circle can be related to the radius of the circle using the formula:
\[r^2 = d_1 \cdot d_2\]
where $r$ is the radius of the circle, and $d_1$ and $d_2$ are the lengths of two intersecting chords.
In our case, we have triangle $ABC$ inscribed in the circle. We can draw the altitudes from points $B$ and $C$ to the opposite sides $AC$ and $AB$, respectively. Let's call the point of intersection of the altitude from $B$ with $AC$ as $D$, and the point of intersection of the altitude from $C$ with $AB$ as $E$.
Now, we can find the lengths of $BD$ and $CE$ using the Pythagorean theorem. Let's consider triangle $BDC$. Since the altitude from $B$ is perpendicular to $AC$, we have a right triangle $BDC$. Using the Pythagorean theorem, we have:
\[BD^2 + CD^2 = BC^2\]
\[BD^2 + (AC - AD)^2 = BC^2\]
\[BD^2 + (12 - AD)^2 = 15^2\]
\[BD^2 + (12 - AD)^2 = 225\]
Similarly, we can use triangle $ACE$ and the Pythagorean theorem to find:
\[CE^2 + (12 - AE)^2 = 8^2\]
\[CE^2 + (12 - AE)^2 = 64\]
Now, to find the radius $r$, we can use the formula $r^2 = BD \cdot CE$ since these two chords intersect inside the circle. Substituting the expressions for $BD$ and $CE$ from the previous calculations, we have:
\[r^2 = (12 - AD)^2 \cdot (12 - AE)^2\]
So, the radius of the circle is given by:
\[r = \sqrt{(12 - AD)^2 \cdot (12 - AE)^2}\]
To find the values of $AD$ and $AE$, we can use the fact that the product of the lengths of the segments of intersecting chords is the same. In this case, $AD \cdot BD = AE \cdot CE$. Substituting the expressions for $BD$ and $CE$ from the previous calculations, we have:
\[AD \cdot (12 - AD) = AE \cdot (12 - AE)\]
Now, we have two equations, one for the radius $r$:
\[r = \sqrt{(12 - AD)^2 \cdot (12 - AE)^2}\]
and one for the product of the lengths of the segments of intersecting chords:
\[AD \cdot (12 - AD) = AE \cdot (12 - AE)\]
We need to solve these equations together to find the value of the radius. You can use algebraic techniques or a graphing calculator to find the value of $r$.