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.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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.

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

hmm what did you get for the distance?
About 7.21
hmm is there a radical form for it?
Sorry...?
well. you used the distance formula to get it.... so... that means you'd have a root of it something like \(\bf distance = \sqrt{\qquad}\)
\[Distance= \sqrt{(-2-4)^2+(-2-2)^2}\]
yeap... I got the same, which in decimal form is about 7.21 so, that's correct \(\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ -2}})\quad % (c,d) &({\color{red}{ 4}}\quad ,&{\color{blue}{ 2}}) \end{array}\qquad % distance value d = \sqrt{6^2+4^2}\implies d=\sqrt{52}\impliedby diameter \\ \quad \\ radius=\cfrac{\sqrt{52}}{2}\)
So the radius would be 3.60
using the midpoint formula, the center is at 1,0 thus \(\large (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ 1}},{\color{blue}{ 0}})\qquad radius={\color{purple}{ \frac{\sqrt{52}}{2}}}\)
Wouldn't the center be (1,1)
?
lemme recheck that
Oops, nevermind, my bad, it is 0
yeah.. is 1,0
shoot.. a bit truncated...lemme redo that
\(\bf (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ 1}},{\color{blue}{ 0}})\qquad radius={\color{purple}{ \frac{\sqrt{52}}{2}}} \\ \quad \\ \quad \\ (x-{\color{brown}{ 1}})^2+(y-{\color{blue}{ 0}})^2=\left( {\color{purple}{ \frac{\sqrt{52}}{2}}}\right)^2 \\ \quad \\ (x-1)^2+y=\cfrac{(\sqrt{52})^2}{2^2}\implies (x-1)^2+y=\cfrac{\cancel{52}}{\cancel{4}} \\ \quad \\ \implies (x-1)^2+y=13\)
So the square root is cancelled off in the equation but I still have a radius of 3.60 right?
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
Yeah, but I need the decimal so that I can graph it
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
welll. yes sure.. but yes, is 3.6 :)
Thank u!!
|dw:1437188896778:dw|
my center is (1, -1)
radius is 1/2 the diameter endpoints (-2,2) and ( 4,2) diameter = 6 units so radius = 3 units for the x value -2 +3 or 4 -3 = 1 from end to the center = 3 units 2 -3 = -1 so the y value for the center = -1
oops I did -2, 2 should be -2, -2 sorry
|dw:1437190476623:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question