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the graph of the equation is a circle. find the center and the radius, …

Question

the graph of the equation is a circle. find the center and the radius, and then sketch.\\(x^{2}+y^{2}+8y = 0\\)\\(\dots\\)use the graphing tool to graph the circle. click to enlarge graph

Explanation:

Step1: Recall circle equation form

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. We need to rewrite the given equation \(x^2 + y^2 + 8y = 0\) into this form by completing the square for the \(y\)-terms.

Step2: Complete the square for \(y\)

For the \(y\)-terms \(y^2 + 8y\), we take half of the coefficient of \(y\) (which is \(8\), so half is \(4\)), square it (\(4^2 = 16\)), and add and subtract it. But since we are only dealing with the \(y\)-terms in the equation \(x^2 + y^2 + 8y = 0\) (the \(x\)-term is already a perfect square: \(x^2=(x - 0)^2\)), we add \(16\) to both sides to complete the square for \(y\):

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Step3: Identify center and radius

Now comparing with the standard form \((x - h)^2 + (y - k)^2 = r^2\), we have \(h = 0\), \(k = - 4\), and \(r^2 = 16\), so \(r = \sqrt{16}=4\).

Answer:

The center of the circle is \((0, -4)\) and the radius is \(4\).