Question

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a certain circle can be represented by the following equation.
$x^2 + y^2 + 4x - 18y + 81 = 0$
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 \(x^{2}+y^{2}+4x - 18y+81 = 0\). Group the \(x\)-terms and \(y\)-terms together: \((x^{2}+4x)+(y^{2}-18y)= - 81\).

Step2: Complete the square for x-terms

For the \(x\)-terms \(x^{2}+4x\), we take half of the coefficient of \(x\) (which is \(4\)), so \(\frac{4}{2}=2\), and square it: \(2^{2} = 4\). Add and subtract this value (but we will just add it to both sides for completing the square). So we have \((x^{2}+4x + 4)+(y^{2}-18y)=-81 + 4\).

Step3: Complete the square for y-terms

For the \(y\)-terms \(y^{2}-18y\), take half of the coefficient of \(y\) (which is \(- 18\)), so \(\frac{-18}{2}=-9\), and square it: \((-9)^{2}=81\). Add this to both sides: \((x^{2}+4x + 4)+(y^{2}-18y + 81)=-81 + 4+81\).

Step4: Write in standard form of circle equation

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. From the completed square, we have \((x + 2)^{2}+(y - 9)^{2}=4\). Which can be written as \((x-(-2))^{2}+(y - 9)^{2}=2^{2}\).

Answer:

The center of the circle is \((-2,9)\) and the radius is \(2\) units.

For the center: \((-2, 9)\)

For the radius: \(2\)