Question

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

Explanation:

Step1: Recall the general equation of a circle

The general equation of a circle is \((x - a)^{2}+(y - b)^{2}=r^{2}\) (for a circle centered at \((a,b)\) with radius \(r\)). In the case of a circle centered at the origin \((0,0)\), the equation is \(x^{2}+y^{2}=r^{2}\). We can also write equations of circles in the form \(Ax^{2}+Ay^{2}=C\) (\(A
eq0\)), which can be simplified to \(x^{2}+y^{2}=\frac{C}{A}\)

Step2: Analyze each option

• Option A: \(\frac{x^{2}}{20}+\frac{y^{2}}{20}=1\) can be rewritten as \(x^{2}+y^{2}=20\). The radius \(r = \sqrt{20}\approx4.47\).
• Option B: Divide \(6x^{2}+6y^{2}=144\) by \(6\). We get \(x^{2}+y^{2}=24\). The radius \(r=\sqrt{24}\approx4.9\).
• Option C: \(20x^{2}-20y^{2}=400\) can be rewritten as \(x^{2}-y^{2}=20\), which is the equation of a hyperbola (\(\frac{x^{2}}{20}-\frac{y^{2}}{20}=1\)), not a circle.
• Option D: Assume the equation is \(2x^{2}+2y^{2}=32\). Divide by \(2\) gives \(x^{2}+y^{2}=16\), radius \(r = 4\)

From the graph, the radius of the circle is approximately \(r=\sqrt{24}\). When we simplify \(6x^{2}+6y^{2}=144\) (divide both sides by \(6\)) to get \(x^{2}+y^{2}=24\), which represents a circle centered at the origin with radius \(r = \sqrt{24}\approx4.9\) (by counting the units on the coordinate - plane, we can see that this is the best fit among the given options).

Answer:

B. \(6x^{2}+6y^{2}=144\)