Graph a circle that has its center at 0,0 and passes through 3,3
Cannot graph on these posts, so I don't know how Helper Girl got the "awencer" of 20.
equation:
x^2 + y^2 = r^2
(3,3) on it, so
9 + 9 = r^2
equation: x^2 + y^2 = 18
graph: (copy and paste into URL)
www.wolframalpha.com/input/?i=plot+x%5E2+%2B+y%5E2+%3D+18
To graph a circle with its center at (0,0) and passing through (3,3), we can use the equation of a circle, which is (x - h)^2 + (y - k)^2 = r^2, where (h,k) represents the center of the circle and r represents the radius.
In this case, the center is (0,0) and one point on the circle is (3,3). We can use these coordinates to find the radius.
Step 1: Find the radius
The distance between the center (0,0) and the point on the circle (3,3) can be found using the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
Substituting the coordinates, we get:
d = sqrt((3 - 0)^2 + (3 - 0)^2) = sqrt(3^2 + 3^2) = sqrt(9 + 9) = sqrt(18) = 3√2.
So, the radius of the circle is 3√2.
Step 2: Plot the circle
Now that we know the radius, we can plot the circle. Since the center is (0,0), we start by plotting this point on the coordinate plane.
Then, we draw a circle with a radius of 3√2 around the center (0,0). To do this, we can plot points at an equal distance from the center, gradually increasing the distance until we reach the radius.
In this case, we can plot the following points:
(0,0), (1,1), (2,2), (3,3), (–1,1), (–2,2), (–3,3), (1,–1), (2,–2), (3,–3), (–1,–1), (–2,–2), and (–3,–3).
Connecting these points will create a circle passing through (3,3) with its center at (0,0).