Question

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

Answer 1

`(x-3)^2+(y-7/2)^2= 85/4`

Explanation:

`r^2=x^2+y^2`
`x=rcostheta`
`y=rsintheta`

`r^2= 6rcostheta+7rsintheta`

`x^2+y^2= 6x+7y`

`x^2-6x+9+y^2-7y+49/4=9+49/4`

`(x-3)^2+(y-7/2)^2= 85/4`

Answer 2

`(x-3)^2+(y-7/2)^2= (sqrt85/2)^2`

Explanation:

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

Multiply both sides of the equation by `r`:

`r^2=6rcos(theta)+7rsin(theta)`

Substitute `r^2 = x^2+y^2`, `y =rsin(theta)`, and `x = rcos(theta)`:

`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^2-6x+y^2-7y=0`

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

`x^2-6x+ h^2+y^2-7y+k^2=h^2+k^2`

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

`-2hx= -6x` and `-2ky = -7y`

`h= 3` and `k = 7/2`

Write the left side as squares and the right side as `3^2+(7/2)^2`:

`(x-3)^2+(y-7/2)^2= 3^2+(7/2)^2`

Simplify the right side:

`(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-7/2)^2= (sqrt85/2)^2`

Answer 3

Rectangular form is `(x -3)^2 +(y-3.5)^2 =21.25`

Explanation:

We know ,`r^2=x^2+y^2 , x= r cos theta , y= r sin theta`

`r = 6 cos theta +7 sin theta` or

`r*r = (6 cos theta +7 sin theta)*r` or

`r^2 = 6 r cos theta +7 r sin theta` or

`x^2+y^2 = 6 x +7 y` or

`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` or

`(x -3)^2 +(y-3.5)^2 =21.25`

Rectangular form is `(x -3)^2 +(y-3.5)^2 =21.25`

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