How do you convert $9 = {(x - 7)}^{2} + {(y + 4)}^{2}$ $9=(x-7)^2+(y+4)^2$ into polar form?

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Answer 1 · 2016-04-18T15:18:37

$r^{2} - 14 r cos θ + 8 r sin θ + 56 = 0$ $r^2-14rcos theta+8rsintheta+56=0$

$9 = x^{2} - 14 x + 49 + y^{2} + 8 y + 16$ $9=x^2-14x+49+y^2+8y+16$ ->FOIL

$x^{2} + y^{2} - 14 x + 8 y + (49 + 16 - 9) = 0$ $x^2+y^2 -14x +8y+(49+16-9)=0$

$x^{2} + y^{2} - 14 x + 8 y + 56 = 0$ $x^2+y^2 -14x +8y+56=0$

Use these formulas:
$r^{2} = x^{2} + y^{2}, x = r cos θ, y = r sin θ$ $r^2=x^2+y^2, x=rcos theta, y=rsin theta$

$r^{2} - 14 r cos θ + 8 r sin θ + 56 = 0$ $r^2-14rcos theta+8rsintheta+56=0$

How do you convert 9=(x-7)^2+(y+4)^29=(x−7)2+(y+4)29=(x-7)^2+(y+4)^2 into polar form?