Question

A circle is centered at N (-6 -2) The point E (-1, 1) is on the circle. Where does the point H (-10, -7) lie?

Answer 1

outside the circle

Explanation:

trust me. i did it on khan academy

Answer 2

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?

`\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 = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ \stackrel{radius}{r}=√([-1-(-6)]^2+[1-(-2)]^2)\implies r=√((-1+6)^2+(1+2)^2) \\\\\\ r=√(5^2+3^2)\implies r=√(34)\implies r\approx 5.83 \\\\[-0.35em] ~\dotfill`

`\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 = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ NH=√([-10-(-6)]^2+[-7-(-2)]^2) \\\\\\ NH=√((-10+6)^2+(-7+2)^2)\implies NH=√((-4)^2+(-5)^2) \\\\\\ NH=√(41)\implies NH\approx 6.4\impliedby \begin{array}{llll} \textit{units away from the center}\\ \textit{is outside the circle} \end{array}`

recall the radius is about 5.83, anything shorter than that is inside the circle, anything longer than that is outside it.