What is the equation of the circle with its center at (0,0) that passes through (3,4).
Given that the center of the circle is the point (0,0).
We know that the equation of the circle is given by :
(x-a)^2 + (y-b)^2 = r^2 where (a,b) is the center and r is the radius.
==> x^2 + y^2 = r^2
Now we will determine the radius.
Since the point (3,4) is on the circle, then the distance between the point and the center is the radius.
==> r = sqrt( 3^2 + 4^2) = sqrt25 = 5
Then, the radius of the circle is 5 units.
==> Them the equation of the circle is:
x^2 + y^2 = 25
The general equation of a circle is (x - a)^2 + (y - b)^2 = r^2, where the center is at (a , b) and the radius is r.
Now we have the center of the circle at (0,0). It passes through the point (3,4).
So the radius is sqrt ( 3^2 + 4^2)
=> sqrt ( 16 + 9)
=> sqrt 25
Substituting the coordinates of the center (0,0) and the radius 5, we get (x - 0)^2 + (y - 0)^2 = 5^2
=> x^2 + y^2 = 25
The equation of the circle is x^2 + y^2 = 25