## Naudiapie one year ago What is the equation of a circle that passes through (2, 0) and has its center at M(4, 0)?

1. Naudiapie

2. anonymous

standard form of circle: (x - h)^2 + (y - k)^2 = r^2center is (h , k) and radius is rfor your circle, the center (h , k) is (1 , 2)so it will look like: (x - 1)^2 + (y - 2)^2 = r^2to find the radius squared, plug in the (x,y) point you know is on the circle -- (3 , 4)(3 - 1)^2 + (4 - 2)^2 = r^22^2 + 2^2 = 8 = r^2

3. jdoe0001

|dw:1437264018410:dw|

4. Naudiapie

???

5. anonymous

Ok, so we know the standard form of a circle is (x - h) ² + (y - k) ² = r ² and the center is (h , k) and radius is r. Does this make sense so far?

6. Naudiapie

yes

7. Naudiapie

!!!

8. jim_thompson5910

How far is it from (4,0) to (2,0) ?

9. Naudiapie

(2,0) ?? i dont know xD

10. jim_thompson5910

|dw:1437263870646:dw|

11. jim_thompson5910

do you see how they are 2 units apart ? |dw:1437263878771:dw|

12. Naudiapie

yes i see that

13. anonymous

I'll brb, I have to finish some of my tests

14. jim_thompson5910

|dw:1437264011320:dw|

15. Naudiapie

16. Naudiapie

it makes no sense to me

17. jim_thompson5910

The general circle equation is this |dw:1437264186407:dw|

18. jim_thompson5910

(h,k) is the center r is the radius

19. jim_thompson5910

we're told that the center is (4,0) so (h,k) = (4,0) meaning that h = 4, k = 0 |dw:1437264245046:dw|

20. jim_thompson5910

|dw:1437264266650:dw|

21. jim_thompson5910

r = 2 was just found and shown above |dw:1437264297767:dw|

22. jim_thompson5910

|dw:1437264320072:dw|

23. jim_thompson5910

also notice how \(\Large (y-0)^2\) turned into \(\Large y^2\)

24. Naudiapie

how did you get 2 squared, can you explain that?

25. Naudiapie

that's what i don't get

26. jim_thompson5910

well the right side of the general circle equation has \(\Large r^2\) I replaced r with 2 because the radius is r = 2

27. jim_thompson5910

r = 2 \[\Large r^2 = 2^2\]

28. jim_thompson5910

|dw:1437264515515:dw|

29. Naudiapie

ohh i see okay

30. Naudiapie

so x - 4squared + ysquared = 2squared (aka 4)

31. jim_thompson5910

yes \[\Large (x-4)^2 + y^2 = 4\]

32. jim_thompson5910

make sure you use parenthesis don't say x-4^2 say (x-4)^2

33. Naudiapie

okay thanks a lot , i get it now

34. jim_thompson5910

4^2 means "4 squared"

35. Naudiapie

thank you lots

36. anonymous

You could also have used the equation of the circle to figure out the radius algebraicly. We knew \[ (x-4)^2+y^2=r^2 \]And we knew \((x,y)=(2,0)\) lie in the circle, so \[ (2-4)^2+0^2=r^2 \\ (-2)^2 = r^2 \implies |r| = 2 \]But since we know \(r\) is positive for it to be a circle, we pick \(r=2\).

37. Naudiapie

:o

38. jim_thompson5910

you're welcome