Question

a certain circle can be represented by the following equation.
$x^2 + y^2 + 12x + 4y + 15 = 0$
what is the center of this circle ?
(\boxed{\quad}, \boxed{\quad})
what is the radius of this circle ?
\boxed{\quad} units

Explanation:

Step1: Rewrite the equation by grouping x and y terms

We start with the equation \(x^{2}+y^{2}+12x + 4y+15 = 0\). Group the \(x\)-terms and \(y\)-terms together: \((x^{2}+12x)+(y^{2}+4y)=- 15\).

Step2: Complete the square for x-terms

For the \(x\)-terms \(x^{2}+12x\), we take half of the coefficient of \(x\) (which is \(\frac{12}{2} = 6\)), square it (\(6^{2}=36\)) and add it to both sides.

Step3: Complete the square for y-terms

For the \(y\)-terms \(y^{2}+4y\), we take half of the coefficient of \(y\) (which is \(\frac{4}{2}=2\)), square it (\(2^{2} = 4\)) and add it to both sides.

After completing the square, the equation becomes:
\((x^{2}+12x + 36)+(y^{2}+4y + 4)=-15 + 36+4\)

Step4: Rewrite as standard circle equation

The left - hand side can be written as perfect squares: \((x + 6)^{2}+(y + 2)^{2}=25\)

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

Comparing \((x + 6)^{2}+(y + 2)^{2}=25\) with \((x - h)^{2}+(y - k)^{2}=r^{2}\), we have \(h=-6\), \(k = - 2\) and \(r^{2}=25\), so \(r = 5\) (since radius is non - negative).

Answer:

The center of the circle is \((-6,-2)\) and the radius is \(5\) units.