Question

consider a circle whose equation is (x^{2}+y^{2}+4x - 6y-36 = 0). which statements are true? check all that apply.
to begin converting the equation to standard form, subtract 36 from both sides.
to complete the square for the x terms, add 4 to both sides.
the center of the circle is at ((-2,3)).
the center of the circle is at ((4,-6)).
the radius of the circle is 6 units.
the radius of the circle is 49 units.

Explanation:

Step1: Rewrite the given equation

Given \(x^{2}+y^{2}+4x - 6y-36 = 0\). To convert to standard form \((x - a)^{2}+(y - b)^{2}=r^{2}\) (where \((a,b)\) is the center and \(r\) is the radius), we first group \(x\) - terms and \(y\) - terms: \((x^{2}+4x)+(y^{2}-6y)=36\). So, we should add \(36\) to both sides (not subtract), so the first statement is false.

Step2: Complete the square for \(x\) - terms

For \(x^{2}+4x\), using the formula \((x + m)^{2}=x^{2}+2mx+m^{2}\). Here \(2m = 4\Rightarrow m = 2\), and \(m^{2}=4\). For \(y^{2}-6y\), using \((y - n)^{2}=y^{2}-2ny + n^{2}\), here \(2n=6\Rightarrow n = 3\), and \(n^{2}=9\).
Adding \(4\) (for \(x\) - terms) and \(9\) (for \(y\) - terms) to both sides of \((x^{2}+4x)+(y^{2}-6y)=36\), we get \((x^{2}+4x + 4)+(y^{2}-6y+9)=36 + 4+9\).

Step3: Simplify to standard form

\((x + 2)^{2}+(y - 3)^{2}=49\). The standard form of a circle is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where the center is \((a,b)\) and radius \(r\). Here \(a=-2\), \(b = 3\), so the center is \((-2,3)\). And \(r^{2}=49\Rightarrow r = 7\).

Answer:

• The second statement (To complete the square for the \(x\) terms, add \(4\) to both sides) is true.
• The third statement (The center of the circle is at \((-2,3)\)) is true.