Find the equation of the circle that satisfies the given conditions.
Center at the origin and passes through (7, 9)
130 = ?
any circle with centre at the origin has equation
x^2 + y^2 = r^2
simply sub in the point(7,9) to find the value of r^2
A circle is centered at the point (5, -4) and passes through the point (-3, 2).
To find the equation of a circle with the center at the origin and passing through a given point, we can use the standard form of the equation of a circle: (x - h)^2 + (y - k)^2 = r^2.
Given that the center is at the origin (h = 0, k = 0), and the circle passes through the point (7, 9), we can substitute these values into the equation:
(7 - 0)^2 + (9 - 0)^2 = r^2
Simplifying this equation gives us:
49 + 81 = r^2
130 = r^2
Therefore, the equation of the circle that satisfies these conditions is x^2 + y^2 = 130.
To find the equation of a circle, we need to know its center (h, k) and its radius (r). In this case, the center of the circle is given as the origin (0, 0), and it passes through the point (7, 9).
The distance between the center of the circle and any point on the circle is equal to the radius. Therefore, we can use the distance formula to find the radius.
The distance formula is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Substituting the given points into the distance formula, we get:
r = √((7 - 0)^2 + (9 - 0)^2)
r = √(49 + 81)
r = √130
Now that we have the radius, we can write the equation of the circle in the standard form:
(x - h)^2 + (y - k)^2 = r^2
Since the center of the circle is at the origin (0, 0) and the radius is √130, the equation of the circle is:
x^2 + y^2 = 130