Question

type the correct answer in each box. use numerals instead of words. if necessary, use / for the fraction bar(s). the equation of a circle is given. $x^{2}+y^{2}+6x + 10y + 18 = 0$ determine the center and radius of the circle. the center of the circle is at (\boxed{ } , \boxed{ }) and the radius of the circle is \boxed{ } units.

Explanation:

Step1: Complete the square for x-terms

For the \(x\) terms \(x^{2}+6x\), we use the formula \((a + b)^2=a^{2}+2ab + b^{2}\). Here, \(a = x\), \(2ab = 6x\) so \(b = 3\). Then \(x^{2}+6x=(x + 3)^{2}-9\).

Step2: Complete the square for y-terms

For the \(y\) terms \(y^{2}+10y\), using the same formula, \(a = y\), \(2ab=10y\) so \(b = 5\). Then \(y^{2}+10y=(y + 5)^{2}-25\).

Step3: Rewrite the circle equation

Substitute the completed squares into the original equation:

$$
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$$

The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. Comparing \((x + 3)^{2}+(y + 5)^{2}=16\) with the standard form, we have \(h=- 3\), \(k = - 5\) and \(r^{2}=16\) so \(r = 4\).

Answer:

The center of the circle is at \((-3, -5)\) and the radius of the circle is \(4\) units.