Question

Graph the equation with a diameter that has endpoints at (-2, -2) and (4, 2).
Label the center and at least four points on the
circle. Write the equation of the circle.

Answer 1

@jdoe0001

Answer 2

Heey, this is kinda the same question as befre, but I don't think my answer for distance is correct

Answer 3

@Directrix

Answer 4

hmm what did you get for the distance?

Answer 5

About 7.21

Answer 6

hmm is there a radical form for it?

Answer 7

Sorry...?

Answer 8

well. you used the distance formula to get it.... so... that means you'd have a root of it
something like \(\bf distance = \sqrt{\qquad}\)

Answer 9

\[Distance= \sqrt{(-2-4)^2+(-2-2)^2}\]

Answer 10

yeap... I got the same, which in decimal form is about 7.21

so, that's correct \(\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({\color{red}{ -2}}\quad ,&{\color{blue}{ -2}})\quad
% (c,d)
&({\color{red}{ 4}}\quad ,&{\color{blue}{ 2}})
\end{array}\qquad
% distance value
d = \sqrt{6^2+4^2}\implies d=\sqrt{52}\impliedby diameter
\\ \quad \\
radius=\cfrac{\sqrt{52}}{2}\)

Answer 11

So the radius would be 3.60

Answer 12

using the midpoint formula, the center is at 1,0

thus \(\large (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2

\qquad center\ ({\color{brown}{ 1}},{\color{blue}{ 0}})\qquad
radius={\color{purple}{ \frac{\sqrt{52}}{2}}}\)

Answer 13

Wouldn't the center be (1,1)

Answer 14

?

Answer 15

lemme recheck that

Answer 16

Oops, nevermind, my bad, it is 0

Answer 17

yeah.. is 1,0

Answer 18

shoot.. a bit truncated...lemme redo that

Answer 19

\(\bf (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2

\qquad center\ ({\color{brown}{ 1}},{\color{blue}{ 0}})\qquad
radius={\color{purple}{ \frac{\sqrt{52}}{2}}}
\\ \quad \\ \quad \\

(x-{\color{brown}{ 1}})^2+(y-{\color{blue}{ 0}})^2=\left( {\color{purple}{ \frac{\sqrt{52}}{2}}}\right)^2
\\ \quad \\
(x-1)^2+y=\cfrac{(\sqrt{52})^2}{2^2}\implies (x-1)^2+y=\cfrac{\cancel{52}}{\cancel{4}}
\\ \quad \\
\implies (x-1)^2+y=13\)

Answer 20

So the square root is cancelled off in the equation but I still have a radius of 3.60 right?

Answer 21

if you want the decimal form of it, yes
the advantage of the radical form is that, it doesn't have any missing floating values

Answer 22

Yeah, but I need the decimal so that I can graph it

Answer 23

but yes, 7.21 is a rounded up version, you lose a few decimals there
half that is 3.6, also rounded up, again losing a few decimals

Answer 24

welll. yes sure.. but yes, is 3.6 :)

Answer 25

Thank u!!