Find Cartesian form of equation r=9cos (theta)?

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Find Cartesian form of equation r=9cos (theta)?

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Answer 1 · 2018-06-26T07:39:14

$x^{2} + y^{2} - 9 x = 0$ $x^2+y^2-9x=0$

Explanation:

we need the rectangular $\to$ $rarr$ Polar transformations

$r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$

$x = r cos θ$ $x=rcostheta$

$y = r sin θ$ $y=rsintheta$

we have

$r = 9 cos θ$ $r=9costheta$

multiply by $r$ $r$

$r^{2} = 9 r cos θ$ $r^2=9rcostheta$

using the transformations above

$x^{2} + y^{2} = 9 x$ $x^2+y^2=9x$

$x^{2} + y^{2} - 9 x = 0$ $x^2+y^2-9x=0$

Answer 2 · 2018-06-26T07:48:11

$x^{2} - 9 x + y^{2} = 0$ $x^2-9x+y^2=0$

Explanation:

$to convert from polar to cartesian$ $"to convert from "color(blue)"polar to cartesian"$

$∙ x r = \sqrt{x^{2} + y^{2}}$ $•color(white)(x)r=sqrt(x^2+y^2)$

$∙ x x = r cos θ \Rightarrow cos θ = \frac{x}{r}$ $•color(white)(x)x=rcosthetarArrcostheta=x/r$

$r = \frac{9 x}{r}$ $r=(9x)/r$

$multiply through by r$ $"multiply through by "r$

$r^{2} = 9 x$ $r^2=9x$

$x^{2} + y^{2} = 9 x$ $x^2+y^2=9x$

$x^{2} - 9 x + y^{2} = 0$ $x^2-9x+y^2=0$

${(x - \frac{9}{2})}^{2} + y^{2} = \frac{81}{4}$ $(x-9/2)^2+y^2=81/4$

$which is the equation of a circle$ $"which is the equation of a circle "$

$centre = (\frac{9}{2}, 0), radius = \frac{9}{2}$ $"centre "=(9/2,0)," radius "=9/2$

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Find Cartesian form of equation r=9cos (theta)?

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