Question

question 13 of 20 which of the following equations will produce the graph shown below? a. $x^{2}+y^{2}=16$ b. $6x^{2}+6y^{2}=144$ c. $\frac{x^{2}}{20}+\frac{y^{2}}{20}=1$ d. $20x^{2}-20y^{2}=400$

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.

Step2: Determine the radius from the graph

From the graph, the circle is centered at the origin \((0,0)\) and intersects the \(x\) - axis and \(y\) - axis at \(x = 4,x=-4,y = 4,y = - 4\). So the radius \(r = 4\). Then the equation of the circle is \(x^{2}+y^{2}=16\) since \(r^{2}=16\).

Step3: Check other equations

For option B: \(6x^{2}+6y^{2}=144\), divide through by 6 to get \(x^{2}+y^{2}=24\), \(r^{2}=24\), \(r=\sqrt{24}
eq4\).
For option C: \(\frac{x^{2}}{20}+\frac{y^{2}}{20}=1\) can be rewritten as \(x^{2}+y^{2}=20\), \(r^{2}=20\), \(r = \sqrt{20}
eq4\).
For option D: \(20x^{2}-20y^{2}=400\) is a hyperbola equation (\(\frac{x^{2}}{20}-\frac{y^{2}}{20}=2\)) and not a circle equation.

Answer:

A. \(x^{2}+y^{2}=16\)