Question

Convert the equation r = 12 sin theta to Cartesian coordinates. Describe the resulting curve.

Answer 1

Answer:

Step-by-step explanation:

Given that there is a polar equation as

$r=12 sin \theta$

This has to be converted into cartesian.

We know the conversion is

$r^2 =x^2+y^2 \\tan \theta = \frac{y}{x}$

Using this we can say that

$sin^2 \theta = 1-cos^2 \theta \\= 1-\frac{1}{sec^2 \theta} \\=1-\frac{1}{1+tan^2 \theta} \\=1-\frac{1}{1+\frac{y^2}{x^2} } \\=1-\frac{x^2}{x^2+y^2} \\=\frac{y^2}{y^2+y^2}$

$\sqrt{x^2+y^2} =12(\frac{y}{\sqrt{x^2+y^2} } \\x^2+y^2 =12y\\x^2+y^2-12y+36 = 36\\x^2+(y-6)^2 = 6^2$}

Circle with centre (0,6) and radius 6.