Question

Convert r = 6 sin [theta] − 2 cos [theta] into rectangular form

Answer 1

Answer:

$x^2+y^2-6x+2y=0$

Step-by-step explanation:

Using the formula to convert polar to rectangular form:

1. $r^2= x^2+y^2$
2. $x= r sin \theta$
3. $y= r cos\theta$

Given that

$r= 6 sin \theta -2 cos\theta$

Multiplying r both sides

$\Rightarrow r^2= 6 rsin \theta -2r cos\theta$

$\Rightarrow x^2+y^2= 6x-2y$

$\Rightarrow x^2+y^2-6x+2y=0$

which is a equation of circle: center (3,-1) and $r=\sqrt{3^2+(-2)^2}=\sqrt{13}$ units.

Answer 2

Answer:

The rectangular form of the r=6sinθ-2cosθ is  x= -1  , y= 3 and r=$\sqrt{10}$, i.e

R(-1,3) and r=$\sqrt{10}$.

Step-by-step explanation:

Given:

Equation in polar form as ; r=6sinθ-2cosθ.

To find ;

Rectangular form of given Polar form .

Solution:

We know that relationship between the polar and rectangular co-ordinates.

Hence,

r=√x²+y²

Sin∅=y/r

Cos∅=x/r

Tan∅=y/x.

Using above relations in given form,

r=6sin∅-2cos∅

r=6(y/r)-2(x/r)

x²+y²=6y-2x

x²+2x+y²-6y=0

This above equation is of the circle

Now calculating the center and radius of circle will be give us the rectangular coordinate for the  given equation.

adding '1' and '9' we get completing square on both sides

x²+2x+y²-6y+1+9=1+9.

(x²+2x+1)+ (y²-6y+9)=10.

(x+1)²+(y-3)²=(√10)².

Hence comparing with standard equation of circle as ,

(x-h)²+(y-k)²=r².

here,

x-h=x+1,

h=-1,

And

y-k=y-3

k=3.

Hence coordinate are (-1,3) with radius √10.