Question

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0​

Answer 1

Radius:
`r =\frac{\sqrt {21}}{6}`

`Center = (-(3)/(2), -(2)/(3))`

Explanation:

Given

`9x^2 + 9y^2 + 27x + 12y + 19 = 0`

Solving (a): The radius of the circle

First, we express the equation as:

`(x - h)^2 + (y - k)^2 = r^2`

Where

`r = radius`

`(h,k) =center`

So, we have:

`9x^2 + 9y^2 + 27x + 12y + 19 = 0`

Divide through by 9

`x^2 + y^2 + 3x + (12)/(9)y + (19)/(9) = 0`

Rewrite as:

`x^2 + 3x + y^2+ (12)/(9)y =- (19)/(9)`

Group the expression into 2

`[x^2 + 3x] + [y^2+ (12)/(9)y] =- (19)/(9)`

`[x^2 + 3x] + [y^2+ (4)/(3)y] =- (19)/(9)`

Next, we complete the square on each group.

For
`[x^2 + 3x]`

1: Divide the
`coefficient\ of\ x\ by\ 2`

2: Take the
`square\ of\ the\ division`

3: Add this
`square\ to\ both\ sides\ of\ the\ equation.`

So, we have:

`[x^2 + 3x] + [y^2+ (4)/(3)y] =- (19)/(9)`

`[x^2 + 3x + ((3)/(2))^2] + [y^2+ (4)/(3)y] =- (19)/(9)+ ((3)/(2))^2`

Factorize

`[x + (3)/(2)]^2+ [y^2+ (4)/(3)y] =- (19)/(9)+ ((3)/(2))^2`

Apply the same to y

`[x + (3)/(2)]^2+ [y^2+ (4)/(3)y +((4)/(6))^2 ] =- (19)/(9)+ ((3)/(2))^2 +((4)/(6))^2`

`[x + (3)/(2)]^2+ [y +(4)/(6)]^2 =- (19)/(9)+ ((3)/(2))^2 +((4)/(6))^2`

`[x + (3)/(2)]^2+ [y +(4)/(6)]^2 =- (19)/(9)+ (9)/(4) +(16)/(36)`

Add the fractions

`[x + (3)/(2)]^2+ [y +(4)/(6)]^2 =(-19 * 4 + 9 * 9 + 16 * 1)/(36)`

`[x + (3)/(2)]^2+ [y +(4)/(6)]^2 =(21)/(36)`

`[x + (3)/(2)]^2+ [y +(4)/(6)]^2 =(7)/(12)`

`[x + (3)/(2)]^2+ [y +(2)/(3)]^2 =(7)/(12)`

Recall that:

`(x - h)^2 + (y - k)^2 = r^2`

By comparison:

`r^2 =(7)/(12)`

Take square roots of both sides

`r =\sqrt{(7)/(12)}`

Split

`r =(\sqrt 7)/(\sqrt 12)`

Rationalize

`r =(\sqrt 7*\sqrt 12)/(\sqrt 12*\sqrt 12)`

`r =\frac{\sqrt {84}}{12}`

`r =\frac{\sqrt {4*21}}{12}`

`r =\frac{2\sqrt {21}}{12}`

`r =\frac{\sqrt {21}}{6}`

Solving (b): The center

Recall that:

`(x - h)^2 + (y - k)^2 = r^2`

Where

`r = radius`

`(h,k) =center`

From:

`[x + (3)/(2)]^2+ [y +(2)/(3)]^2 =(7)/(12)`

`-h = (3)/(2)` and
`-k = (2)/(3)`

Solve for h and k

`h = -(3)/(2)` and
`k = -(2)/(3)`

Hence, the center is:

`Center = (-(3)/(2), -(2)/(3))`