Question

I don't know if this is right... please someone help mee

Answer 1

For the first circle, let's use the pythagorean theorem

$\bf \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17$

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

$\bf \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938$

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

$\bf \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}$

this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.