Respuesta :

Answer:

[tex]r= 2(cos(\theta) + 3sin(\theta))[/tex]

Step-by-step explanation:

To convert an equation of rectangular shape to polar form use the following equivalences

[tex]x=rcos(\theta)[/tex]

[tex]y=rsin(\theta)[/tex]

[tex]x^2 + y^2=r^2[/tex]

In this case we have the following equation.

[tex](x-1)^2 + (y-3)^2=10[/tex]

First we expanded the expression

[tex](x-1)^2 + (y-3)^2=10\\\\(x^2 -2x +1) +(y^2 -6y + 9) = 10\\\\\\x^2 +y^2 -2x -6y = 10-9-1\\\\x^2 +y^2 -2x -6y = 0[/tex]

Now we know that [tex]x^2 + y^2=r^2[/tex], [tex]x=rcos(\theta)[/tex] and [tex]y=rsin(\theta)[/tex]

So

[tex]r^2 -2rcos(\theta) -6rsin(\theta) = 0\\\\r^2 - 2r(cos(\theta) + 3sin(\theta))=0\\\\r^2 = 2r(cos(\theta) + 3sin(\theta))\\\\r= 2(cos(\theta) + 3sin(\theta))[/tex]