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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. x-coordinate of center: y-coordinate of center: Radius:

Mathematics
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What is ab\ne 0?
r=a sin (theta) + b cos (theta), ab is not = 0
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?

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Other answers:

yea
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
where does a and b come in?
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?
Yeah sure you can do that.
this can give r^2= ay + bx right?
Ok and r^2=x^2+y^2
yes
so that will give x^2+y^2=ay+bx
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
so meaning (x-b)^2+(y-a)^2
?
hmm....I don't you completed the square right
I mean you didn't complete the square...
come again
\[x^2-bx+y^2-ay=0\] you must complete the square honestly you made some rules up or something
ok
can u help with that?
\[x^2+cx+(\frac{c}{2})^2=(x+\frac{c}{2})^2\]
why (c/2)^2 and not (cx/2)^2?
\[(x+\frac{c}{2})^2=(x+\frac{c}{2})(x+\frac{c}{2})=x(x+\frac{c}{2})+\frac{c}{2}(x+\frac{c}{2})\] \[=x^2+\frac{c}{2}x+\frac{c}{2}x+(\frac{c}{2})^2=x^2+cx+(\frac{c}{2})^2\]
ok
what next?
\[x^2-bx+y^2-ay=0 \]
You need to complete the square for both the x part and the y part.
so we get (x+c/2)^2+(y+b/2)^2=0?
then what next? equate each part to 0?
hmmm.... So you have x^2-bx+b/2=(x-b/2)^2
But whatever you add to one side you add to the other .
x=b/2,y =a/2 and r=sqrt(a^2+b^2)/2
YEP that is right.
thank you

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