Question

a circle has the equation 2x² + 24x + 2y² = 0. (a) find the center (h,k) and radius r of the circle. (b) graph the circle. (c) find the intercepts, if any, of the graph. (a) the center of the circle is (-6,0). (type an ordered pair, using integers or decimals.) the radius of the circle is 6. (type an integer or a decimal.) (b) use the graphing tool to graph the circle. (c) what are the intercepts? select the correct choice and, if necessary, fill in the answer box within your choice. a. the intercept(s) of the circle is/are . (type an ordered pair. use a comma to separate answers as needed. type exact answers for each coordinate, using radicals as needed.) b. there are no intercepts.

Explanation:

Step1: Recall the standard - form of a circle equation

The general equation of a circle is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius. Given the equation \(2x^{2}+24x + 2y^{2}=0\), first divide through by 2 to get \(x^{2}+12x + y^{2}=0\). Then complete the square for the \(x\) - terms: \(x^{2}+12x=(x + 6)^{2}-36\). So the equation becomes \((x + 6)^{2}-36+y^{2}=0\), or \((x + 6)^{2}+y^{2}=36\). The center \((h,k)=(-6,0)\) and radius \(r = 6\).

Step2: Find the \(x\) - intercepts

Set \(y = 0\) in the equation \((x + 6)^{2}+y^{2}=36\). Then \((x + 6)^{2}=36\). Taking the square - root of both sides, we have \(x+6=\pm6\). When \(x + 6 = 6\), \(x=0\); when \(x + 6=-6\), \(x=-12\).

Step3: Find the \(y\) - intercepts

Set \(x = 0\) in the equation \((x + 6)^{2}+y^{2}=36\). Then \((0 + 6)^{2}+y^{2}=36\), which simplifies to \(36+y^{2}=36\), so \(y = 0\).

Answer:

A. \((-12,0),(0,0)\)