Change r=6cos(theta)+7sin(theta) to rectangular form ?

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Change r=6cos(theta)+7sin(theta) to rectangular form ?

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Answer 1 · 2018-05-15T12:22:38

${(x - 3)}^{2} + {(y - \frac{7}{2})}^{2} = \frac{85}{4}$ $(x-3)^2+(y-7/2)^2= 85/4$

Explanation:

$r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$
$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

$r^{2} = 6 r cos θ + 7 r sin θ$ $r^2= 6rcostheta+7rsintheta$

$x^{2} + y^{2} = 6 x + 7 y$ $x^2+y^2= 6x+7y$

$x^{2} - 6 x + 9 + y^{2} - 7 y + \frac{49}{4} = 9 + \frac{49}{4}$ $x^2-6x+9+y^2-7y+49/4=9+49/4$

${(x - 3)}^{2} + {(y - \frac{7}{2})}^{2} = \frac{85}{4}$ $(x-3)^2+(y-7/2)^2= 85/4$

Answer 2 · 2018-05-15T12:44:32

${(x - 3)}^{2} + {(y - \frac{7}{2})}^{2} = {(\frac{\sqrt{85}}{2})}^{2}$ $(x-3)^2+(y-7/2)^2= (sqrt85/2)^2$

Explanation:

Given: $r = 6 cos (θ) + 7 sin (θ)$ $r=6cos(theta)+7sin(theta)$

Multiply both sides of the equation by $r$ $r$ :

$r^{2} = 6 r cos (θ) + 7 r sin (θ)$ $r^2=6rcos(theta)+7rsin(theta)$

Substitute $r^{2} = x^{2} + y^{2}$ $r^2 = x^2+y^2$ , $y = r sin (θ)$ $y =rsin(theta)$ , and $x = r cos (θ)$ $x = rcos(theta)$ :

$x^{2} + y^{2} = 6 x + 7 y$ $x^2+y^2=6x+7y$

Technically, we are done but we recognize that the equation is not in a standard form, therefore, we shall proceed.

Move everything to the left so that the equation equals 0:

$x^{2} - 6 x + y^{2} - 7 y = 0$ $x^2-6x+y^2-7y=0$

Add $h^{2} + k^{2}$ $h^2+k^2$ to both sides so that we can complete the squares:

$x^{2} - 6 x + h^{2} + y^{2} - 7 y + k^{2} = h^{2} + k^{2}$ $x^2-6x+ h^2+y^2-7y+k^2=h^2+k^2$

Use the middle terms to find the values of $h$ $h$ and $k$ $k$ :

$- 2 h x = - 6 x$ $-2hx= -6x$ and $- 2 k y = - 7 y$ $-2ky = -7y$

$h = 3$ $h= 3$ and $k = \frac{7}{2}$ $k = 7/2$

Write the left side as squares and the right side as $3^{2} + {(\frac{7}{2})}^{2}$ $3^2+(7/2)^2$ :

${(x - 3)}^{2} + {(y - \frac{7}{2})}^{2} = 3^{2} + {(\frac{7}{2})}^{2}$ $(x-3)^2+(y-7/2)^2= 3^2+(7/2)^2$

Simplify the right side:

${(x - 3)}^{2} + {(y - \frac{7}{2})}^{2} = \frac{85}{4}$ $(x-3)^2+(y-7/2)^2= 85/4$

To comply with the standard Cartesian form of the equation of a circle, we should write the right side as a square:

${(x - 3)}^{2} + {(y - \frac{7}{2})}^{2} = {(\frac{\sqrt{85}}{2})}^{2}$ $(x-3)^2+(y-7/2)^2= (sqrt85/2)^2$

Answer 3 · 2018-05-15T12:48:16

Rectangular form is ${(x - 3)}^{2} + {(y - 3.5)}^{2} = 21.25$ $(x -3)^2 +(y-3.5)^2 =21.25$

Explanation:

We know , $r^{2} = x^{2} + y^{2}, x = r cos θ, y = r sin θ$ $r^2=x^2+y^2 , x= r cos theta , y= r sin theta$

$r = 6 cos θ + 7 sin θ$ $r = 6 cos theta +7 sin theta$ or

$r \cdot r = (6 cos θ + 7 sin θ) \cdot r$ $r*r = (6 cos theta +7 sin theta)*r$ or

$r^{2} = 6 r cos θ + 7 r sin θ$ $r^2 = 6 r cos theta +7 r sin theta$ or

$x^{2} + y^{2} = 6 x + 7 y$ $x^2+y^2 = 6 x +7 y$ or

$x^{2} - 6 x + y^{2} - 7 y = 0$ $x^2 -6 x +y^2 -7 y =0$ or

$x^{2} - 6 x + 9 + y^{2} - 7 y + {3.5}^{2} = 9 + 12.25$ $x^2 -6 x +9 +y^2 -7 y +3.5^2=9 +12.25$ or

${(x - 3)}^{2} + {(y - 3.5)}^{2} = 21.25$ $(x -3)^2 +(y-3.5)^2 =21.25$

Rectangular form is ${(x - 3)}^{2} + {(y - 3.5)}^{2} = 21.25$ $(x -3)^2 +(y-3.5)^2 =21.25$

graph{x^2+y^2= 6 x+7 y [-20, 20, -10, 10]} [Ans]

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Change r=6cos(theta)+7sin(theta) to rectangular form ?

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