Question

a certain circle can be represented by the following equation.
$x^2 + y^2 - 4x + 12y - 24 = 0$
what is the center of this circle ?
( , )
what is the radius of this circle ?
units

Explanation:

Step1: Rewrite the equation by grouping x and y terms

We start with the equation of the circle: \(x^{2}+y^{2}-4x + 12y-24=0\). Group the \(x\)-terms and \(y\)-terms together: \((x^{2}-4x)+(y^{2}+12y)=24\).

Step2: Complete the square for x-terms

For the \(x\)-terms \(x^{2}-4x\), we use the formula \((a - b)^2=a^{2}-2ab + b^{2}\). Here, \(a = x\) and \(2ab=4x\), so \(b = 2\). Then \(x^{2}-4x=(x - 2)^{2}-4\).

Step3: Complete the square for y-terms

For the \(y\)-terms \(y^{2}+12y\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), where \(a = y\) and \(2ab = 12y\), so \(b=6\). Then \(y^{2}+12y=(y + 6)^{2}-36\).

Step4: Substitute back and simplify

Substitute the completed square forms back into the equation: \((x - 2)^{2}-4+(y + 6)^{2}-36=24\). Simplify the left - hand side: \((x - 2)^{2}+(y + 6)^{2}-40=24\). Then \((x - 2)^{2}+(y + 6)^{2}=24 + 40=64\).

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.

Answer:

The center of the circle is \((2,-6)\) and the radius is \(8\) units.