How do you convert #9=(x-6)^2+(y+2)^2# into polar form?

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Answer 1 · 2018-07-25T16:57:43

#r^2 - 4sqrt10 r cos ( theta -tan^(-1)(-1/3)) + 31 = 0#

The center of this circle is #C ( 6, -2 ) in Q_4#.

Radius of the circle = 3 and.

Radius from pole ( O), #OC = c = sqrt ( 6^2 + (- 2 )^2) = 2sqrt10#.

The inclination of the polar radius ( from O ) OC

#alpha= tan^(-1)( y_C/x_C)#

# = tan^( - 1 ) (- 1 / 3 ) in Q_4#. So,

polar coordinates of C are ( c, alpha ) = ( 2 sqrt10, alpha )#..

If P ( r, Theta ) is any point on the circle, CP =3 and

#CP^2 = 9 = OP^2 + OC^2 - 2(OC)(OP).cos ( theta - alpha )#

#= r^2 + 40 - 4 sqrt 10 r cos ( theta - alpha )# giving

#r^2 - 4 sqrt 10 r cos ( theta -tan^(-1)(-1/3)) + 31 = 0#

The general equation is of the form

#r^2 - 2 r c cos (theta - alpha ) + c^2 - a^2 = 0#..