Respuesta :
so we know the point E is on the circle, thus the distance NE is really the radius of the circle hmmm what would that be?
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-2})\qquad E(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[-1-(-6)]^2+[1-(-2)]^2}\implies r=\sqrt{(-1+6)^2+(1+2)^2} \\\\\\ r=\sqrt{5^2+3^2}\implies r=\sqrt{34}\implies r\approx 5.83 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-2})\qquad H(\stackrel{x_2}{-10}~,~\stackrel{y_2}{-7})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NH=\sqrt{[-10-(-6)]^2+[-7-(-2)]^2} \\\\\\ NH=\sqrt{(-10+6)^2+(-7+2)^2}\implies NH=\sqrt{(-4)^2+(-5)^2} \\\\\\ NH=\sqrt{41}\implies NH\approx 6.4\impliedby \begin{array}{llll} \textit{units away from the center}\\ \textit{is outside the circle} \end{array}[/tex]
recall the radius is about 5.83, anything shorter than that is inside the circle, anything longer than that is outside it.
Answer:
outside the circle
Step-by-step explanation:
trust me. i did it on khan academy
