anonymous one year ago 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.

1. anonymous

@jdoe0001

2. anonymous

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

3. anonymous

@Directrix

4. anonymous

hmm what did you get for the distance?

5. anonymous

6. anonymous

hmm is there a radical form for it?

7. anonymous

Sorry...?

8. anonymous

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

9. anonymous

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

10. anonymous

yeap... I got the same, which in decimal form is about 7.21 so, that's correct $$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ -2}})\quad % (c,d) &({\color{red}{ 4}}\quad ,&{\color{blue}{ 2}}) \end{array}\qquad % distance value d = \sqrt{6^2+4^2}\implies d=\sqrt{52}\impliedby diameter \\ \quad \\ radius=\cfrac{\sqrt{52}}{2}$$

11. anonymous

So the radius would be 3.60

12. anonymous

using the midpoint formula, the center is at 1,0 thus $$\large (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ 1}},{\color{blue}{ 0}})\qquad radius={\color{purple}{ \frac{\sqrt{52}}{2}}}$$

13. anonymous

Wouldn't the center be (1,1)

14. anonymous

?

15. anonymous

lemme recheck that

16. anonymous

Oops, nevermind, my bad, it is 0

17. anonymous

yeah.. is 1,0

18. anonymous

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

19. anonymous

$$\bf (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ 1}},{\color{blue}{ 0}})\qquad radius={\color{purple}{ \frac{\sqrt{52}}{2}}} \\ \quad \\ \quad \\ (x-{\color{brown}{ 1}})^2+(y-{\color{blue}{ 0}})^2=\left( {\color{purple}{ \frac{\sqrt{52}}{2}}}\right)^2 \\ \quad \\ (x-1)^2+y=\cfrac{(\sqrt{52})^2}{2^2}\implies (x-1)^2+y=\cfrac{\cancel{52}}{\cancel{4}} \\ \quad \\ \implies (x-1)^2+y=13$$

20. anonymous

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

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

Thank u!!

26. triciaal

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27. triciaal

my center is (1, -1)

28. 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

29. triciaal

oops I did -2, 2 should be -2, -2 sorry

30. triciaal

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