How do you convert $(x-3)^2+(y+1)^2=10$ to polar form?

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Answer 1 · 2016-03-10T23:15:27

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

First use the binomial theorem to express the squared terms as polynomials. Recall that $(a+b)^2=a^2+2ab+b^2$ . So

$(x-3)^2=x^2-6x+9$
$(y+1)^2=y^2+2y+1$

Now write the left side ss the sum of these polynomials. You should put $x^2+y^2$ together because we know that in polar form that is $r^2$ .

$(x^2+y^2)-6x+2y+10=10$

Now put in the coordinate conversions

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

Then

$r^2-6r$ cos $\theta+2r$ sin $\theta=0$

This can be factored so we have two possibilities:

$r=0$

$r-6$ cos $\theta+2$ sin $\theta=0$

The first equation is just the origin and the second equation already has a point at the origin. So we just use the second equation.

How do you convert (x-3)^2+(y+1)^2=10(x-3)^2+(y+1)^2=10 to polar form?