How do you convert ${(x - 3)}^{2} + {(y + 4)}^{2} = 25$ $(x-3)^2+(y+4)^2=25$ into polar form?

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Answer 1 · 2016-11-19T02:05:42

Please see the explanation for steps leading to:

$r = 6 cos (θ) - 8 sin (θ)$ $r = 6cos(theta) - 8sin(theta)$

Expand the squares:

$x^{2} - 6 x + 9 + y^{2} + 8 y + 16 = 25$ $x^2 - 6x + 9 + y^2 + 8y + 16 = 25$

Group the square terms, the linear terms and please notice that the constants sum to zero:

$x^{2} + y^{2} - 6 x + 8 y = 0$ $x^2 + y^2 - 6x + 8y = 0$

Substitute $r^{2}$ $r^2$ for $x^{2} + y^{2}$ $x^2 + y^2$

$r^{2} - 6 x + 8 y = 0$ $r^2 - 6x + 8y = 0$

Substitute $r cos (θ)$ $rcos(theta)$ for x and $r sin (θ)$ $rsin(theta)$ for y:

$r^{2} - 6 r cos (θ) + 8 r sin (θ) = 0$ $r^2 - 6rcos(theta) + 8rsin(theta) = 0$

Discard a common r factor:

$r - 6 cos (θ) + 8 sin (θ) = 0$ $r - 6cos(theta) + 8sin(theta) = 0$

Move the sine and cosine terms to the right side:

$r = 6 cos (θ) - 8 sin (θ)$ $r = 6cos(theta) - 8sin(theta)$

How do you convert (x-3)^2+(y+4)^2=25(x−3)2+(y+4)2=25(x-3)^2+(y+4)^2=25 into polar form?