Question

assignment 10.0: distance and midpoint formulas; circles
score: 78/100 answered: 10/12
question 11
$3x^{2}+3y^{2}-18x - 30y - 6 = 0$ is the equation of a circle with center $(h,k)$ and radius $r$ for:
$h=$
and
$k=$
and
$r=$
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Explanation:

Step1: Divide the equation by 3

Divide each term in \(3x^{2}+3y^{2}-18x - 30y-6 = 0\) by 3.
We get \(x^{2}+y^{2}-6x - 10y-2 = 0\).

Step2: Complete the square for x - terms

For the \(x\) - terms \(x^{2}-6x\), we use the formula \((a - b)^2=a^{2}-2ab + b^{2}\). Here \(a=x\) and \(2ab = 6x\), so \(b = 3\). Then \(x^{2}-6x=(x - 3)^{2}-9\).

Step3: Complete the square for y - terms

For the \(y\) - terms \(y^{2}-10y\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\). Here \(a = y\) and \(2ab=10y\), so \(b = 5\). Then \(y^{2}-10y=(y - 5)^{2}-25\).

Step4: Rewrite the equation

Substitute the completed - square expressions into \(x^{2}+y^{2}-6x - 10y-2 = 0\).
\((x - 3)^{2}-9+(y - 5)^{2}-25-2 = 0\).
Simplify to \((x - 3)^{2}+(y - 5)^{2}=36\).

Answer:

\(h = 3\), \(k = 5\), \(r = 6\)