## anonymous 3 years ago It can be shown that the polar curve r=a\sin(\theta)+b\cos(\theta), where ab\ne 0, is a circle. Find the center (in Cartesian coordinates) and radius of this circle in terms of a and b. x-coordinate of center: y-coordinate of center: Radius:

1. anonymous

What is ab\ne 0?

2. anonymous

r=a sin (theta) + b cos (theta), ab is not = 0

3. anonymous

Do you agree, that in cartesian coordinates the radius is $\sqrt{(x-x_0)^2+(y-y_0)^2}$Where (x0, y0) is the center?

4. anonymous

yea

5. myininaya

You could write it as a Cartesian equation by using cos(theta)=x/r and sin(theta)=y/r and r^2=x^2+y^2

6. anonymous

where does a and b come in?

7. anonymous

i know that multiplying the equation by r is a legitimate step that can get me r^2=a rsin (theta) + b rcos (theta). Is that right?

8. myininaya

Yeah sure you can do that.

9. anonymous

this can give r^2= ay + bx right?

10. myininaya

Ok and r^2=x^2+y^2

11. anonymous

yes

12. anonymous

so that will give x^2+y^2=ay+bx

13. myininaya

ok and write that equation in this form: (x-h)^2+(y-k)^2=r^2 where (h,k) is center and r is radius

14. anonymous

so meaning (x-b)^2+(y-a)^2

15. anonymous

?

16. myininaya

hmm....I don't you completed the square right

17. myininaya

I mean you didn't complete the square...

18. anonymous

come again

19. myininaya

$x^2-bx+y^2-ay=0$ you must complete the square honestly you made some rules up or something

20. anonymous

ok

21. anonymous

can u help with that?

22. myininaya

$x^2+cx+(\frac{c}{2})^2=(x+\frac{c}{2})^2$

23. anonymous

why (c/2)^2 and not (cx/2)^2?

24. myininaya

$(x+\frac{c}{2})^2=(x+\frac{c}{2})(x+\frac{c}{2})=x(x+\frac{c}{2})+\frac{c}{2}(x+\frac{c}{2})$ $=x^2+\frac{c}{2}x+\frac{c}{2}x+(\frac{c}{2})^2=x^2+cx+(\frac{c}{2})^2$

25. anonymous

ok

26. anonymous

what next?

27. myininaya

$x^2-bx+y^2-ay=0$

28. myininaya

You need to complete the square for both the x part and the y part.

29. anonymous

so we get (x+c/2)^2+(y+b/2)^2=0?

30. anonymous

then what next? equate each part to 0?

31. myininaya

hmmm.... So you have x^2-bx+b/2=(x-b/2)^2

32. myininaya

But whatever you add to one side you add to the other .

33. anonymous

x=b/2,y =a/2 and r=sqrt(a^2+b^2)/2

34. myininaya

YEP that is right.

35. anonymous

thank you