Question

burtin, jacameron
geometry c&c unit 6 pre-test
the equation below describes a circle.
\\(x^2 + 6x + y^2 - 10y = 31\\)
what are the center and radius of the circle?
\\(\bigcirc\\) the center is \\((3, -5)\\) and the radius is \\(\sqrt{65}\\).
\\(\bigcirc\\) the center is \\((-3, 5)\\) and the radius is \\(\sqrt{65}\\).
\\(\bigcirc\\) the center is \\((3, -5)\\) and the radius is \\(\sqrt{31}\\).
\\(\bigcirc\\) the center is \\((-3, 5)\\) and the radius is \\(\sqrt{31}\\).

Explanation:

Step1: Complete the square for x

The equation is \(x^{2}+6x + y^{2}-10y = 31\). For the \(x\)-terms, take the coefficient of \(x\), which is 6, divide by 2 to get 3, and square it to get 9. So we add 9 to both sides for the \(x\)-part: \(x^{2}+6x + 9 + y^{2}-10y=31 + 9\). This factors to \((x + 3)^{2}+y^{2}-10y=40\).

Step2: Complete the square for y

For the \(y\)-terms, the coefficient of \(y\) is -10, divide by 2 to get -5, square it to get 25. Add 25 to both sides: \((x + 3)^{2}+y^{2}-10y + 25=40 + 25\). This factors to \((x + 3)^{2}+(y - 5)^{2}=65\).

Step3: Identify center and radius

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 \((x + 3)^{2}+(y - 5)^{2}=65\) with the standard form, we have \(h=-3\), \(k = 5\), and \(r^{2}=65\), so \(r=\sqrt{65}\).

Answer:

The center is \((-3,5)\) and the radius is \(\sqrt{65}\). (Corresponding to the option: The center is \((-3,5)\) and the radius is \(\sqrt{65}\).)