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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.
xcoordinate of center:
ycoordinate of center:
Radius:
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. xcoordinate of center: ycoordinate of center: Radius:

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0r=a sin (theta) + b cos (theta), ab is not = 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you agree, that in cartesian coordinates the radius is \[\sqrt{(xx_0)^2+(yy_0)^2}\]Where (x0, y0) is the center?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0You 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where does a and b come in?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah sure you can do that.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this can give r^2= ay + bx right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so that will give x^2+y^2=ay+bx

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0ok and write that equation in this form: (xh)^2+(yk)^2=r^2 where (h,k) is center and r is radius

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so meaning (xb)^2+(ya)^2

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0hmm....I don't you completed the square right

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0I mean you didn't complete the square...

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0\[x^2bx+y^2ay=0\] you must complete the square honestly you made some rules up or something

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can u help with that?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0\[x^2+cx+(\frac{c}{2})^2=(x+\frac{c}{2})^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why (c/2)^2 and not (cx/2)^2?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0\[(x+\frac{c}{2})^2=(x+\frac{c}{2})(x+\frac{c}{2})=x(x+\frac{c}{2})+\frac{c}{2}(x+\frac{c}{2})\] \[=x^2+\frac{c}{2}x+\frac{c}{2}x+(\frac{c}{2})^2=x^2+cx+(\frac{c}{2})^2\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0You need to complete the square for both the x part and the y part.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so we get (x+c/2)^2+(y+b/2)^2=0?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then what next? equate each part to 0?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0hmmm.... So you have x^2bx+b/2=(xb/2)^2

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0But whatever you add to one side you add to the other .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x=b/2,y =a/2 and r=sqrt(a^2+b^2)/2
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