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

Answer:

Radius: [tex]r =\frac{\sqrt {21}}{6}[/tex]

[tex]Center = (-\frac{3}{2}, -\frac{2}{3})[/tex]

Step-by-step explanation:

Given

[tex]9x^2 + 9y^2 + 27x + 12y + 19 = 0[/tex]

Solving (a): The radius of the circle

First, we express the equation as:

[tex](x - h)^2 + (y - k)^2 = r^2[/tex]

Where

[tex]r = radius[/tex]

[tex](h,k) =center[/tex]

So, we have:

[tex]9x^2 + 9y^2 + 27x + 12y + 19 = 0[/tex]

Divide through by 9

[tex]x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0[/tex]

Rewrite as:

[tex]x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}[/tex]

Group the expression into 2

[tex][x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}[/tex]

[tex][x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}[/tex]

Next, we complete the square on each group.

For [tex][x^2 + 3x][/tex]

1: Divide the [tex]coefficient\ of\ x\ by\ 2[/tex]

2: Take the [tex]square\ of\ the\ division[/tex]

3: Add this [tex]square\ to\ both\ sides\ of\ the\ equation.[/tex]

So, we have:

[tex][x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}[/tex]

[tex][x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2[/tex]

Factorize

[tex][x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2[/tex]

Apply the same to y

[tex][x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2[/tex]

[tex][x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2[/tex]

[tex][x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}[/tex]

Add the fractions

[tex][x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}[/tex]

[tex][x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}[/tex]

[tex][x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}[/tex]

[tex][x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}[/tex]

Recall that:

[tex](x - h)^2 + (y - k)^2 = r^2[/tex]

By comparison:

[tex]r^2 =\frac{7}{12}[/tex]

Take square roots of both sides

[tex]r =\sqrt{\frac{7}{12}}[/tex]

Split

[tex]r =\frac{\sqrt 7}{\sqrt 12}[/tex]

Rationalize

[tex]r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}[/tex]

[tex]r =\frac{\sqrt {84}}{12}[/tex]

[tex]r =\frac{\sqrt {4*21}}{12}[/tex]

[tex]r =\frac{2\sqrt {21}}{12}[/tex]

[tex]r =\frac{\sqrt {21}}{6}[/tex]

Solving (b): The center

Recall that:

[tex](x - h)^2 + (y - k)^2 = r^2[/tex]

Where

[tex]r = radius[/tex]

[tex](h,k) =center[/tex]

From:

[tex][x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}[/tex]

[tex]-h = \frac{3}{2}[/tex] and [tex]-k = \frac{2}{3}[/tex]

Solve for h and k

[tex]h = -\frac{3}{2}[/tex] and [tex]k = -\frac{2}{3}[/tex]

Hence, the center is:

[tex]Center = (-\frac{3}{2}, -\frac{2}{3})[/tex]

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