anonymous
  • anonymous
Challenge: Graph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the circle. Write the equation of the circle.
Mathematics
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anonymous
  • anonymous
Challenge: Graph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the circle. Write the equation of the circle.
Mathematics
chestercat
  • chestercat
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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.

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jdoe0001
  • jdoe0001
hmm what's the distance between (-3, 4) and (5, -2). ? |dw:1437177936311:dw|
anonymous
  • anonymous
eq.of circle is (x-x1)(x-x2)+(y-y1)(y-y2)=0
anonymous
  • anonymous
About 9.90

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anonymous
  • anonymous
So I divide that by two to get the radius right? And with it the center
jdoe0001
  • jdoe0001
@sewandowski hmmm how did you get 9.9?
anonymous
  • anonymous
Square root of: (-3-4)^2 + (5-(-2))^2
anonymous
  • anonymous
mid point is the center
anonymous
  • anonymous
Oooh, so I use the midpoint formula?
anonymous
  • anonymous
yes
jdoe0001
  • jdoe0001
hmm anyhow as @surjithayer said find the diameter to get the "radius" ahd the midpoint formula to get the center
jdoe0001
  • jdoe0001
btw, the distance aint' 9.9 either
anonymous
  • anonymous
Ok, so the midpoint of the points is (1,1), which is the center
jdoe0001
  • jdoe0001
yeap...1,1 is the center
jdoe0001
  • jdoe0001
and the diameter?
anonymous
  • anonymous
Ok, now I got 10
anonymous
  • anonymous
so the radius is 5
jdoe0001
  • jdoe0001
yeap radius is 5, diameter is 10 then \(\large (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad radius={\color{purple}{ r}}\)
anonymous
  • anonymous
Perfect, now all I need to do is graph
anonymous
  • anonymous
Thanks guys!!
jdoe0001
  • jdoe0001
yw
anonymous
  • anonymous
eq. of circle is \[\left\{ x-\left( -3 \right) \right\}\left( x-5 \right)+\left( y-4 \right)\left\{ y-\left( -2 \right) \right\}=0\] solve it.

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