Find an equation of the circle with center at the origin and passing through (2,-5) in the form of (x-A)^2+(y-B)^2=C where A,B,C are constant.
a circle centered at (h,k) with radius r has the general equation
... (x - h)^2 + (y - k)^2 = r^2
in this case ; h and k are zero (origin) , and ... r^2 = 2^2 + (-5)^2
To find the equation of the circle with center at the origin and passing through a given point, we can use the general equation of a circle:
(x - A)^2 + (y - B)^2 = r^2
where A and B are the coordinates of the center and r is the radius of the circle.
In this case, since the center of the circle is at the origin, A = 0 and B = 0. Therefore, the equation becomes:
x^2 + y^2 = r^2
We still need to find the value of r, which is the radius of the circle.
Since the circle passes through the point (2, -5), we can substitute these coordinates into the equation:
(2)^2 + (-5)^2 = r^2
4 + 25 = r^2
29 = r^2
Now we have the value of r^2, which is 29. So the equation of the circle is:
x^2 + y^2 = 29