anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@jdoe0001
anonymous
  • anonymous
Heey, this is kinda the same question as befre, but I don't think my answer for distance is correct
anonymous
  • anonymous
@Directrix

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jdoe0001
  • jdoe0001
hmm what did you get for the distance?
anonymous
  • anonymous
About 7.21
jdoe0001
  • jdoe0001
hmm is there a radical form for it?
anonymous
  • anonymous
Sorry...?
jdoe0001
  • jdoe0001
well. you used the distance formula to get it.... so... that means you'd have a root of it something like \(\bf distance = \sqrt{\qquad}\)
anonymous
  • anonymous
\[Distance= \sqrt{(-2-4)^2+(-2-2)^2}\]
jdoe0001
  • jdoe0001
yeap... I got the same, which in decimal form is about 7.21 so, that's correct \(\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ -2}})\quad % (c,d) &({\color{red}{ 4}}\quad ,&{\color{blue}{ 2}}) \end{array}\qquad % distance value d = \sqrt{6^2+4^2}\implies d=\sqrt{52}\impliedby diameter \\ \quad \\ radius=\cfrac{\sqrt{52}}{2}\)
anonymous
  • anonymous
So the radius would be 3.60
jdoe0001
  • jdoe0001
using the midpoint formula, the center is at 1,0 thus \(\large (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ 1}},{\color{blue}{ 0}})\qquad radius={\color{purple}{ \frac{\sqrt{52}}{2}}}\)
anonymous
  • anonymous
Wouldn't the center be (1,1)
anonymous
  • anonymous
?
jdoe0001
  • jdoe0001
lemme recheck that
anonymous
  • anonymous
Oops, nevermind, my bad, it is 0
jdoe0001
  • jdoe0001
yeah.. is 1,0
jdoe0001
  • jdoe0001
shoot.. a bit truncated...lemme redo that
jdoe0001
  • jdoe0001
\(\bf (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ 1}},{\color{blue}{ 0}})\qquad radius={\color{purple}{ \frac{\sqrt{52}}{2}}} \\ \quad \\ \quad \\ (x-{\color{brown}{ 1}})^2+(y-{\color{blue}{ 0}})^2=\left( {\color{purple}{ \frac{\sqrt{52}}{2}}}\right)^2 \\ \quad \\ (x-1)^2+y=\cfrac{(\sqrt{52})^2}{2^2}\implies (x-1)^2+y=\cfrac{\cancel{52}}{\cancel{4}} \\ \quad \\ \implies (x-1)^2+y=13\)
anonymous
  • anonymous
So the square root is cancelled off in the equation but I still have a radius of 3.60 right?
jdoe0001
  • jdoe0001
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
anonymous
  • anonymous
Yeah, but I need the decimal so that I can graph it
jdoe0001
  • jdoe0001
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
jdoe0001
  • jdoe0001
welll. yes sure.. but yes, is 3.6 :)
anonymous
  • anonymous
Thank u!!
triciaal
  • triciaal
|dw:1437188896778:dw|
triciaal
  • triciaal
my center is (1, -1)
triciaal
  • triciaal
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
triciaal
  • triciaal
oops I did -2, 2 should be -2, -2 sorry
triciaal
  • triciaal
|dw:1437190476623:dw|

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