Question

give the center and radius of the circle represented by the equation.
$x^{2}+y^{2}-12x + 6y=-20$
the center of the circle is $square$. (type an ordered pair, using integers or fractions.)
the radius of the circle is $square$. (type an integer or a simplified fraction.)

Explanation:

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

Group \(x\) - terms and \(y\) - terms: \((x^{2}-12x)+(y^{2}+6y)=-20\).
For \(x\) - terms: \(x^{2}-12x=(x - 6)^{2}-36\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a = x\), \(b = 6\)).

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 \(a = y\), \(b = 3\)).

Step3: Substitute back into the equation

Substitute into the original equation: \((x - 6)^{2}-36+(y + 3)^{2}-9=-20\).
Simplify: \((x - 6)^{2}+(y + 3)^{2}=-20 + 36+9\).

Step4: Calculate the right - hand side

\(-20 + 36+9=25\). So the equation of the circle is \((x - 6)^{2}+(y + 3)^{2}=25\).

The standard form of a circle's equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.

Answer:

The center of the circle is \((6,-3)\).
The radius of the circle is \(5\).