Question

the graph shows the circle with the equation ( x^2 + y^2 = 1 ). describe the types of symmetry that you observe. (select all that apply.)
□ x-axis symmetry
□ y-axis symmetry
□ origin symmetry
□ symmetry about the line ( x = 1 )

Explanation:

Step1: Recall symmetry tests

To check for x - axis symmetry, replace \(y\) with \(-y\) in the equation \(x^{2}+y^{2}=1\). We get \(x^{2}+(-y)^{2}=x^{2}+y^{2} = 1\), which is the same as the original equation. So, the graph is symmetric about the x - axis.

Step2: Check y - axis symmetry

To check for y - axis symmetry, replace \(x\) with \(-x\) in the equation \(x^{2}+y^{2}=1\). We get \((-x)^{2}+y^{2}=x^{2}+y^{2}=1\), which is the same as the original equation. So, the graph is symmetric about the y - axis.

Step3: Check origin symmetry

To check for origin symmetry, replace \(x\) with \(-x\) and \(y\) with \(-y\) in the equation \(x^{2}+y^{2}=1\). We get \((-x)^{2}+(-y)^{2}=x^{2}+y^{2}=1\), which is the same as the original equation. So, the graph is symmetric about the origin.

Step4: Check symmetry about \(x = 1\)

Take a point on the circle, say \((1,0)\). The reflection of \((1,0)\) about the line \(x = 1\) is \((1,0)\) itself. Now take another point \((0,1)\). The reflection of \((0,1)\) about the line \(x=1\) is \((2,1)\). Substitute \(x = 2\) and \(y = 1\) into the equation \(x^{2}+y^{2}\): \(2^{2}+1^{2}=4 + 1=5
eq1\), so \((2,1)\) is not on the circle. Thus, the graph is not symmetric about the line \(x = 1\).

Answer:

x - axis symmetry, y - axis symmetry, origin symmetry