## anonymous one year ago A circle has a center at (8,2). The point (3,7) is on the circle. What is the area of the circle to the nearest tenth of a square unit?

1. triciaal

|dw:1442806632519:dw||dw:1442806745433:dw|

2. anonymous

o.o

3. anonymous

|dw:1442806834199:dw|

4. anonymous

sqrt(8+3)^2 +(2-7)^2

5. anonymous

-*

6. anonymous

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ d= \sqrt{(8-3)^2+(2-7)^2} \\ d= \sqrt{(5)^2+(5)^2} \\ d=~?$

7. anonymous

sqrt25+25 ?

8. anonymous

9. anonymous

Yes and what is 25 + 25?

10. anonymous

50

11. anonymous

Yes, so $$r=\sqrt{50}$$

12. anonymous

$\large A_{\text{circle}} = \pi r^2 = \pi (\sqrt{50})^2 = ~?$

13. anonymous

7

14. anonymous

7?

15. anonymous

sqrt50 =7?

16. anonymous

Okay, yeah I see what you mean

17. anonymous

OOOHHHHH I was putting 50 instead of 7! ;v; Thanks!!

18. anonymous

Typo.

19. anonymous

3.14*7^2=157.1 which is the right answer

20. anonymous

May I ask you another question?

21. anonymous

Well when we square a square root we get the stuff inside the square root :c $(\sqrt{x})^2 = x$ the square root and the square cancel eachother out. So.... I'm not sure where to go from that point.

22. anonymous

Oh I guess you first simplified $$\sqrt{50} \approx 7.0$$ then squared it.

23. anonymous

Yup, may I ask another q?

24. anonymous

I've got to head off OpenStudy right now, but @triciaal can help you! they are very smart as well :) Good luck!

25. anonymous

@triciaal

26. triciaal

this is like going the other way you have r from the area then use the formula I gave above for the circle to find (x, y) you are given (h,k) the center

27. anonymous

Yes, thank you for the formula, but I can't really read all of the text you gave before .-.

28. triciaal

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29. anonymous

25?

30. triciaal

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31. triciaal

leave as r^2

32. anonymous

What are h and k?

33. triciaal

|dw:1442808245627:dw|

34. triciaal

coordinates for center of circle

35. anonymous

Rrriiigghhttt.

36. triciaal

|dw:1442808479367:dw|

37. anonymous

I'm sorry, I really can't understand...

38. anonymous

@triciaal ?

39. triciaal

I don't know how to make you understand I am totally confused now because this is even shown in the diagram I don't know what question you have

40. anonymous

41. anonymous

(-6, 4) (2, -1) (-4, -4) (2, -9)

42. anonymous

43. anonymous

@triciaal ?

44. triciaal

that works

45. anonymous

Yes, but how?

46. anonymous

??