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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:
 one year ago
 one year 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:
 one year ago
 one year ago

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SelormBest ResponseYou've already chosen the best response.0
r=a sin (theta) + b cos (theta), ab is not = 0
 one year ago

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

myininayaBest ResponseYou've already chosen the best response.0
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
 one year ago

SelormBest ResponseYou've already chosen the best response.0
where does a and b come in?
 one year ago

SelormBest ResponseYou've already chosen the best response.0
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?
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
Yeah sure you can do that.
 one year ago

SelormBest ResponseYou've already chosen the best response.0
this can give r^2= ay + bx right?
 one year ago

SelormBest ResponseYou've already chosen the best response.0
so that will give x^2+y^2=ay+bx
 one year ago

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

SelormBest ResponseYou've already chosen the best response.0
so meaning (xb)^2+(ya)^2
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
hmm....I don't you completed the square right
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
I mean you didn't complete the square...
 one year ago

myininayaBest 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
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
\[x^2+cx+(\frac{c}{2})^2=(x+\frac{c}{2})^2\]
 one year ago

SelormBest ResponseYou've already chosen the best response.0
why (c/2)^2 and not (cx/2)^2?
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
\[(x+\frac{c}{2})^2=(x+\frac{c}{2})(x+\frac{c}{2})=x(x+\frac{c}{2})+\frac{c}{2}(x+\frac{c}{2})\] \[=x^2+\frac{c}{2}x+\frac{c}{2}x+(\frac{c}{2})^2=x^2+cx+(\frac{c}{2})^2\]
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
\[x^2bx+y^2ay=0 \]
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
You need to complete the square for both the x part and the y part.
 one year ago

SelormBest ResponseYou've already chosen the best response.0
so we get (x+c/2)^2+(y+b/2)^2=0?
 one year ago

SelormBest ResponseYou've already chosen the best response.0
then what next? equate each part to 0?
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
hmmm.... So you have x^2bx+b/2=(xb/2)^2
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
But whatever you add to one side you add to the other .
 one year ago

SelormBest ResponseYou've already chosen the best response.0
x=b/2,y =a/2 and r=sqrt(a^2+b^2)/2
 one year ago
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