Circle in cartesian and polar

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April 18th 2012, 04:05 PM
Bowlbase

Circle in cartesian and polar

I have a circle whose center is at point (5,-2) and goes through (1,1)

Circle formula: $(x-h)^2+(y-k)^2=r^2$
h=5
k=-2
$r=\sqrt((x_1-x_2)^2+y_1-y_2)^2)$
$r=\sqrt((5-1)^2+(-2-1)^2)$
$r=\sqrt(25)=5$
so,
$(x-5)^2+(y+2)^2=25$

I can't tell for sure if this passes through (1,1) or not from the wolfram graph but looks pretty close.

For polar I'm not sure how to approach this so this is my attempt:
$(x-5)^2+(y+2)^2=25$ = $x^2-10x+25+y^2+4x+4=5=25$
= $x^2-10x+y^2+4x=5=25-29$
making the substitutions for x and y:
= $(rcos(\theta))^2-10rcos(\theta)+(rsin(\theta))^2+4rsin(\theta)=-4$

not sure what to do here on out

thanks for help
April 18th 2012, 05:33 PM
Soroban

Re: Circle in cartesian and polar

Hello, Bowlbase!

Quote:

I have a circle whose center is at point (5,-2) and goes through (1,1)

Circle formula: $(x-h)^2+(y-k)^2\:=\:R^2$
. . $h\,=\,5,\;k\,=\,-2.\;R = 5$

So: . $(x-5)^2+(y+2)^2\:=\:25$

I can't tell for sure if this passes through (1,1) or not from the wolfram graph but looks pretty close.
Substitute x = 1 and y = 1 into the equation . . . Do they satisfy the equation?

To convert to polar coordinates, I'd do it like this . . .

$\begin{array}{c}(r\cos\theta - h)^2 + (r\sin\theta - k)^2 \;=\;25 \\ \\ r^2\cos^2\!\theta - 2rh\cos\theta + h^2 + r^2\sin^2\!\theta - 2rk\sin\theta + k^2 \;=\;25 \\ \\ r^2\underbrace{(\cos^2\!\theta + \sin^2\!\theta)}_{\text{This is 1}} - 2r(h\cos\theta + k\sin\theta) + h^2 + k^2 \;=\;25 \\ \\ r^2 - 2(h\cos\theta + k\sin\theta)r + (h^2+k^2- 25) \;=\;0 \end{array}$

You can use the Quadratic Formula to solve for $r$ . . . but it's pretty ugly!
April 18th 2012, 06:34 PM
Bowlbase

Re: Circle in cartesian and polar

OK, I didn't even think of putting in sin/cos before expanding. That makes it more straight forward than what I was trying to do. Especially with the substitution of h,k last.. it makes it easier to follow.

That helped a lot so thanks a bunch!