for the first circle, let's use the pythagorean theorem
[tex]\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17[/tex]
now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.
so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.
when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.
now, let's check for second circle
[tex]\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938[/tex]
well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.
let's check the third circle
[tex]\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}[/tex]
this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
