Question

Given the circle with the equation (x + 1)2 + y2 = 36, determine the location of each point with respect to the graph of the circle. In your final answer, state whether each point is on the interior, exterior, or circumference of the circle. Include your calculations as proof of each point’s location. A. (-1, 1) B. (-1, 6) C. (4, -8)

Answer 1

Answer:

Step-by-step explanation:

First, we need to plug these points into the equation given to find our inequality. If the answer is lower then the number the equation equals (36) it's on the interior, if it's equal, it's on the circumference, and if it's greater its on the exterior.

A. (-1 + 1)^2 + 1^2 =? 36

0^2 + 1 =? 36

0 + 1 =? 36

We get the value

1 < 36

This point is on the interior of the circle

B. (-1 + 1)^2 + 6^2 =? 36

0^2 + 36 =?36

0 + 36 =? 36

we get the value

36 = 36

This point is on the circumference of the circle

C. (4 + 1)^2 + -8^2 =? 36

5^2 + 64 =? 36

25 + 64 =? 36

we get the value

89 > 36

this point is on the exterior of the circle.

Answer 2

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r}\\\\ -------------------------------\\\\ (x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~ \begin{cases} \stackrel{center}{(-1,0)}\\ \stackrel{radius}{6} \end{cases}$

so, that's the equation of the circle, and that's its center, any point "ON" the circle, namely on its circumference, will have a distance to the center of 6 units, since that's the radius.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2} \\\\\\ d=\sqrt{0+1}\implies d=1$

well, the distance from the center to A is 1, namely is "inside the circle".

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{6})\\\\\\ \stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(6-0)^2}\implies d=\sqrt{(-1+1)^2+6^2} \\\\\\ d=\sqrt{0+36}\implies d=6$

notice, the distance to B is exactly 6, and you know what that means.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-8}) \\\\\\ \stackrel{distance}{d}=\sqrt{[4-(-1)]^2+[-8-0]^2}\implies d=\sqrt{(4+1)^2+(-8)^2} \\\\\\ d=\sqrt{25+64}\implies d=\sqrt{89}\implies d\approx 9.43398$

notice, C is farther than the radius 6, meaning is outside the circle, hiking about on the plane.