How can you prove that the equation of any circle with radius (r) and the center at the origin is x^2+y^2=36?
Please help!?!
YOu cant. The equation you gave is for radius 6 only.
Sorry, I meant how can you prove that the equation of any circle with radius (r) and the center at the origin is x^2+y^2=r^2?
find the distance from any point on the circle to the origin.
The distance from any point x1,y1 to a point x2,y2 is
distance=sqrt ((x1-x2)^2+(y1-y2)^2)
if the distance from any point is constant, that curve is a circle with radius=distance.
To prove that the equation of any circle with radius (r) and center at the origin is x^2 + y^2 = r^2, you need to understand the properties of a circle and its equation.
A circle is a set of points that are equidistant (at the same distance) from a central point called the center. The radius is the distance from the center to any point on the circle.
Now, let's start by proving the equation x^2 + y^2 = r^2 for a circle with radius (r) and the center at the origin (0,0).
The distance between any point (x, y) on the circle and the origin (0, 0) is given by the distance formula as follows:
Distance = √((x - 0)^2 + (y - 0)^2)
= √(x^2 + y^2)
Since all the points on the circle are equidistant from the origin, the distance is equal to the radius (r). Therefore, we can equate the distance to the radius:
√(x^2 + y^2) = r
To eliminate the square root, we square both sides of the equation:
(x^2 + y^2) = r^2
Thus, the equation x^2 + y^2 = r^2 is the equation of any circle with radius (r) and center at the origin (0,0).
Using the specific case you provided with r = 6, we get:
x^2 + y^2 = 6^2
x^2 + y^2 = 36
Therefore, the equation of a circle with a radius of 6 and center at the origin is x^2 + y^2 = 36.