anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
What is ab\ne 0?
anonymous
  • anonymous
r=a sin (theta) + b cos (theta), ab is not = 0
anonymous
  • 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?

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

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