Questions: Here is the equation of a circle: x^2 + y^2 + 12x + 2y + 12 = 0 Find the center and radius of the circle.

Solution Steps

To find the center and radius of the circle given by the equation \(x^{2}+y^{2}+12x+2y+12=0\), we need to rewrite the equation in the standard form of a circle, \((x-h)^2 + (y-k)^2 = r^2\), by completing the square for both \(x\) and \(y\).

The given equation of the circle is:

\[ x^2 + y^2 + 12x + 2y + 12 = 0 \]

To find the center and radius, we need to rewrite this equation in the standard form of a circle's equation:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center and \(r\) is the radius.

First, focus on the \(x\) terms: \(x^2 + 12x\).

To complete the square, take half of the coefficient of \(x\), square it, and add and subtract it inside the equation:

\[ x^2 + 12x = (x^2 + 12x + 36) - 36 = (x + 6)^2 - 36 \]

Next, focus on the \(y\) terms: \(y^2 + 2y\).

Similarly, take half of the coefficient of \(y\), square it, and add and subtract it:

\[ y^2 + 2y = (y^2 + 2y + 1) - 1 = (y + 1)^2 - 1 \]

Substitute the completed squares back into the original equation:

\[ (x + 6)^2 - 36 + (y + 1)^2 - 1 + 12 = 0 \]

Simplify the equation:

\[ (x + 6)^2 + (y + 1)^2 - 25 = 0 \]

Add 25 to both sides to isolate the squared terms:

\[ (x + 6)^2 + (y + 1)^2 = 25 \]

Final Answer