Question

follow the guided instructions below to rotate the figure $90^{circ}$ counter - clockwise about the origin.
draw a circle centered at the center of rotation, such that one of the vertices of the figure is on
the circle.

Explanation:

Step1: Identify a vertex

Let's take the vertex on the y - axis: (0, -2) (wait, looking at the graph, one vertex is (0, -2)? Wait, no, the vertices: let's see, the triangle has vertices, say, (0, -2), (0, -8), and (4, -4) maybe? Wait, the center of rotation is the origin (0,0). Let's take the vertex (0, -2) first. Wait, no, maybe (0, -8) and (4, -4) and (0, -2)? Wait, actually, let's pick a vertex, say (0, -8). The distance from the origin (0,0) to (0, -8) is 8 units (since distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, here $x_1 = 0,y_1 = 0,x_2 = 0,y_2=-8$, so $d=\sqrt{0 + (-8 - 0)^2}=8$).

Step2: Draw the circle

We draw a circle centered at (0,0) with radius equal to the distance from the origin to the chosen vertex. So if we choose the vertex (0, -8), the radius $r = 8$. The equation of the circle is $x^{2}+y^{2}=8^{2}=64$. This circle will pass through the vertex (0, -8) (since when $x = 0,y=-8$, $0+(-8)^{2}=64$, which satisfies the equation). Also, if we choose another vertex, say (4, -4), the distance from origin is $\sqrt{(4 - 0)^2+(-4 - 0)^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}$, so the circle centered at origin with radius $4\sqrt{2}$ will pass through (4, -4). But let's just follow the instruction: draw a circle centered at the center of rotation (origin) such that one of the vertices is on the circle. So we can pick any vertex, calculate its distance from the origin, and draw a circle with that radius centered at origin. For example, take the vertex (0, -2) (wait, no, looking at the graph, the lower vertex is at (0, -8), middle on y - axis at (0, -2)? Wait, maybe the vertices are (0, -2), (4, -4), (0, -8). Let's check (0, -2): distance from origin is 2, (0, -8): distance 8, (4, -4): distance $\sqrt{16 + 16}=\sqrt{32}$. So we can draw a circle centered at (0,0) with radius 8 (passing through (0, -8)), or radius 2 (passing through (0, -2)), or radius $4\sqrt{2}$ (passing through (4, -4)).

Answer:

To draw the circle:

1. Choose a vertex of the triangle (e.g., the vertex at (0, -8)).
2. Calculate the distance from the origin (0,0) to this vertex: using the distance formula $d=\sqrt{(x - 0)^2+(y - 0)^2}$, for (0, -8), $d = \sqrt{0+(-8)^2}=8$.
3. Draw a circle centered at (0,0) with radius 8 (the equation of the circle is $x^{2}+y^{2}=64$) which will pass through the chosen vertex (0, -8).

(Note: If a different vertex is chosen, the radius will change accordingly. For example, for vertex (4, -4), radius is $4\sqrt{2}$ and circle equation is $x^{2}+y^{2}=32$; for vertex (0, -2), radius is 2 and circle equation is $x^{2}+y^{2}=4$)