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

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Answer 1 · 2018-07-25T16:57:43

$r^{2} - 4 \sqrt{10} r cos (θ - {tan}^{- 1} (- \frac{1}{3})) + 31 = 0$ $r^2 - 4sqrt10 r cos ( theta -tan^(-1)(-1/3)) + 31 = 0$

The center of this circle is $C (6, - 2) \in Q_{4}$ $C ( 6, -2 ) in Q_4$ .

Radius of the circle = 3 and.

Radius from pole ( O), $O C = c = \sqrt{6^{2} + {(- 2)}^{2}} = 2 \sqrt{10}$ $OC = c = sqrt ( 6^2 + (- 2 )^2) = 2sqrt10$ .

The inclination of the polar radius ( from O ) OC

$α = {tan}^{- 1} (\frac{y_{C}}{x_{C}})$ $alpha= tan^(-1)( y_C/x_C)$

$= {tan}^{- 1} (- \frac{1}{3}) \in Q_{4}$ $= tan^( - 1 ) (- 1 / 3 ) in Q_4$ . So,

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

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

$C P^{2} = 9 = O P^{2} + O C^{2} - 2 (O C) (O P) . cos (θ - α)$ $CP^2 = 9 = OP^2 + OC^2 - 2(OC)(OP).cos ( theta - alpha )$

$= r^{2} + 40 - 4 \sqrt{10} r cos (θ - α)$ $= r^2 + 40 - 4 sqrt 10 r cos ( theta - alpha )$ giving

$r^{2} - 4 \sqrt{10} r cos (θ - {tan}^{- 1} (- \frac{1}{3})) + 31 = 0$ $r^2 - 4 sqrt 10 r cos ( theta -tan^(-1)(-1/3)) + 31 = 0$

The general equation is of the form

$r^{2} - 2 r c cos (θ - α) + c^{2} - a^{2} = 0$ $r^2 - 2 r c cos (theta - alpha ) + c^2 - a^2 = 0$ ..

How do you convert 9=(x-6)^2+(y+2)^29=(x−6)2+(y+2)29=(x-6)^2+(y+2)^2 into polar form?