Question

5.6.5 test (cst): conic sections which of the following equations will produce the graph shown below? a. 2x² + 2y² = 32 b. 20x² - 20y² = 400 c. $\frac{x^{2}}{20}+\frac{y^{2}}{20}=1$ d. 6x² + 6y² = 144

Explanation:

Step1: Recall the standard form of a circle equation

The standard - form of a circle centered at the origin \((0,0)\) is \(x^{2}+y^{2}=r^{2}\), where \(r\) is the radius of the circle. From the graph, the radius \(r = 4\) (since the circle intersects the \(y\) - axis at \(y = 4\) and \(y=-4\) and the \(x\) - axis at \(x = 4\) and \(x=-4\)), so the equation of the circle should be \(x^{2}+y^{2}=16\).

Step2: Rewrite each option in the form of \(x^{2}+y^{2}=r^{2}\)

Option A:

Given \(2x^{2}+2y^{2}=32\). Divide both sides of the equation by 2: \(\frac{2x^{2}}{2}+\frac{2y^{2}}{2}=\frac{32}{2}\), which simplifies to \(x^{2}+y^{2}=16\).

Option B:

The equation \(20x^{2}-20y^{2}=400\) is a hyperbola equation (of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1\)), not a circle equation.

Option C:

The equation \(\frac{x^{2}}{20}+\frac{y^{2}}{20}=1\) can be rewritten as \(x^{2}+y^{2}=20\), and \(r=\sqrt{20}
eq4\).

Option D:

Given \(6x^{2}+6y^{2}=144\). Divide both sides by 6: \(\frac{6x^{2}}{6}+\frac{6y^{2}}{6}=\frac{144}{6}\), which simplifies to \(x^{2}+y^{2}=24\), and \(r = \sqrt{24}
eq4\).

Answer:

A. \(2x^{2}+2y^{2}=32\)