How do you convert ${(x - 3)}^{2} + {(y - 2)}^{2} = 13$ $(x-3)^2+(y-2)^2=13$ to polar form?

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Answer 1 · 2016-06-22T14:46:20

$r = 3 cos θ + 2 sin θ$ $r=3 cos theta + 2 sin theta$ .

Use the transformation $(x, y) = (r cos θ, r sin θ)$ $(x, y)=(r cos theta, r sin theta)$ .

The given equation reduces to the polar form

$r^{2} - 2 r (3 cos θ + 2 sin θ) = 0$ $r^2-2r(3 cos theta + 2 sin theta)= 0$ .

So, $r = 0 and r = 2 (3 cos θ + 2 sin θ)$ $r = 0 and r=2(3 cos theta + 2 sin theta)$ . The second equation

includes the first at $(0, a r c tan (- \frac{3}{2}))$ $(0, arc tan (-3/2))$ , obtained by solving

. $3 cos θ + 2 sin θ = 0$ $3 cos theta + 2 sin theta=0$ .

The circle passes through the pole.

The radius of the circle is $\sqrt{13}$ $sqrt 13$ .

Note that arc tan (-3/2) has two values = ${123.6}^{o} and {303.69}^{o}$ $123.6^o and 303.69^o$ ,

nearly, in $(0, 360^{o})$ $(0, 360^o)$ , giving the opposite directions of the tangent

to the circle at the origin (pole)..

The center of the circle is at $(\sqrt{13}, a r c tan (- \frac{3}{2}) - or + 90^{o})$ $(sqrt 13, arc tan (-3/2)-or+90^o)$ .

I have taken the trouble of giving some details relating to the polar

form. I think that Just giving the polar equation alone, sans further

details, is an incomplete exercise..

How do you convert (x-3)^2+(y-2)^2=13(x−3)2+(y−2)2=13(x-3)^2+(y-2)^2=13 to polar form?