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=√(a^2+b^2) \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=√(8^2+15^2)\implies c=√(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=√(a^2+b^2) \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=√(11^2+14^2)\implies c=√(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=√(a^2+b^2) \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=√(33^2+56^2)\implies c=√(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.