Question

find the center and the radius of the circle.
$x^{2}+y^{2}+4x + 6y-36 = 0$
center:
radius:

Explanation:

Step1: Rearrange and complete the square for \(x\) terms

Group \(x\) terms and \(y\) terms: \((x^{2}+4x)+(y^{2}+6y)=36\).
For \(x\) terms: \(x^{2}+4x=(x + 2)^{2}-4\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\), \(2b = 4\Rightarrow b = 2\)).

Step2: Complete the square for \(y\) terms

For \(y\) terms: \(y^{2}+6y=(y + 3)^{2}-9\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=y\), \(2b=6\Rightarrow b = 3\)).
Substitute back: \((x + 2)^{2}-4+(y + 3)^{2}-9=36\).

Step3: Write in standard circle form

Simplify to \((x + 2)^{2}+(y + 3)^{2}=36 + 4+9\).
So \((x + 2)^{2}+(y + 3)^{2}=49\).
The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
Here \(h=-2\), \(k=-3\), \(r^{2}=49\Rightarrow r = 7\).

Answer:

Center: \((-2,-3)\)
Radius: \(7\)