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Mathematics 18 Online
OpenStudy (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:

OpenStudy (anonymous):

What is ab\ne 0?

OpenStudy (anonymous):

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

OpenStudy (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?

OpenStudy (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

OpenStudy (anonymous):

where does a and b come in?

OpenStudy (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.

OpenStudy (anonymous):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

yes

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

?

myininaya (myininaya):

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

myininaya (myininaya):

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

OpenStudy (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

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

can u help with that?

myininaya (myininaya):

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

OpenStudy (anonymous):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

ok

OpenStudy (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.

OpenStudy (anonymous):

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

OpenStudy (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 .

OpenStudy (anonymous):

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

myininaya (myininaya):

YEP that is right.

OpenStudy (anonymous):

thank you

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