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

Question

I need to convert (x-1)^2 + (y-3)^2 =10 to polar form and am stuck. I know x=rcos theta and y = rsin theta, but can't make it work out.

Answer 1

Hi Lamar,

You first expand the rectangular equation,

(x-1)^2 + (y-3)^2 = 10

x^2 - 2x + 1 + y^2 - 6y + 9 = 10

simplying and rearranging,

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

replacing (x^2 + y^2) by r^2

replacing x by r cos (theta), y by r sin (theta)

r^2 -2rcos(theta) -6rsin(theta) = 0

divide both sides by r, and rearranging,

r = 2cos(theta) + 6sin(theta)

Please let me know if you have further questions.

Answer 2

In vector form R = (10^(1/2)*cos(a) + 1)i + (10^(1/2)*sin(a) + 3)j where a = theta

or

x = 10^(1/2)*cos(a) + 1 and y = 10^(1/2)*sin(a) + 3

putting back into the equation, you get

((10^(1/2)*cos(a))^2 + ((10^(1/2)*sin(a))^2 = 10