Question

Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necessary. Question 9 options: (x + 3)2 + (y + 6)2 = 2 (x + 3)2 + (y + 6)2 = 8 (x – 3)2 + (y – 6)2 = 2 (x – 3)2 + (y – 6)2 = 8

Answer 1

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2$

Answer 2

Answer:

FIRST OPTION: $(x+3)^2+ (y+6)^2 =2$

Step-by-step explanation:

The equation of the circle in center-radius form is:

$(x- h)^2 + (y- k)^2 = r^2$

Where the center is at the point (h, k) and the radius is "r".

We know that the endpoints of the diameter of this circle are (-4, -7) and (-2, -5), so we can find the radius and the center of the circle.

In order to find the radius, we need to find the diameter. To do this, we need to use the formula for calculate the distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Then, substituting the coordinates of the endpoints of the diameter into this formula, we get:

$d=\sqrt{(-4-(-2))^2+(-7-(-5))^2}=2\sqrt{2}$

Since the radius is half the diameter, this is:

$r=\frac{2\sqrt{2}}{2}=\sqrt{2}$

To find the center, given the endpoints of the diameter, we need to find the midpoint with this formula:

$M=(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})$

This is:

$M=(\frac{-4-2}{2},\frac{-7-5}{2})=(-3,-6)$

Then:

$h=-3\\k=-6$

Finally, substituting values  into  $(x- h)^2 + (y- k)^2 = r^2$, we get the following equation:

 $(x- (-3))^2 + (y- (-6))^2 = (\sqrt{2})^2$

$(x+3)^2+ (y+6)^2 =2$