Question

question
the general form for the equation of a circle is shown.
3x² + 3y² + 30x - 24y - 12 = 0
what is the standard form of the equation?
type the correct answer in the box. use numerals instead of words.
(x + )² + (y + )² =

Explanation:

Step1: Divide the equation by 3

Divide each term in \(3x^{2}+3y^{2}+30x - 24y-12 = 0\) by 3.
$$x^{2}+y^{2}+10x - 8y-4 = 0$$

Step2: Rearrange the terms

Group the \(x\) - terms and \(y\) - terms together.
$$(x^{2}+10x)+(y^{2}-8y)=4$$

Step3: Complete the square for the \(x\) - terms

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

Step4: Complete the square for the \(y\) - terms

For \(y^{2}-8y\), use the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\). Here \(a = y\) and \(2ab=8y\), so \(b = 4\). Then \(y^{2}-8y=(y - 4)^{2}-16\).

Step5: Substitute the completed - square expressions back into the equation

Substitute \((x + 5)^{2}-25\) for \(x^{2}+10x\) and \((y - 4)^{2}-16\) for \(y^{2}-8y\) in \((x^{2}+10x)+(y^{2}-8y)=4\).
$$(x + 5)^{2}-25+(y - 4)^{2}-16=4$$

Step6: Simplify the equation

Combine the constant terms on the right - hand side.
$$(x + 5)^{2}+(y - 4)^{2}=4 + 25+16$$
$$(x + 5)^{2}+(y - 4)^{2}=45$$

Answer:

\((x+5)^{2}+(y - 4)^{2}=45\)