Determine the equation of the circle with center (0,−1) containing the point (−√35,0)
The standard equation for a circle with center (h,k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2
In this case, we know that the center is (0,-1). Therefore, we have:
(x - 0)^2 + (y + 1)^2 = r^2
We also know that the circle contains the point (-√35,0). We can substitute these values into our equation to solve for the radius, r:
(-√35 - 0)^2 + (0 + 1)^2 = r^2
35 + 1 = r^2
r = √36
r = 6
So our equation becomes:
x^2 + (y + 1)^2 = 36
Therefore, the equation of the circle with center (0,-1) and containing the point (-√35,0) is x^2 + (y + 1)^2 = 36.
To determine the equation of the circle with center (0,-1) containing the point (-√35, 0), we can use the standard form equation of a circle:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) is the center of the circle and r is the radius.
Given that the center is (0, -1), we have:
(x - 0)^2 + (y - (-1))^2 = r^2
x^2 + (y + 1)^2 = r^2
To find the radius, we can use the distance formula between the center (0, -1) and the point (-√35, 0):
r^2 = (-√35 - 0)^2 + (0 -(-1))^2
r^2 = 35 + 1
r^2 = 36
r = 6
Therefore, the equation of the circle is:
x^2 + (y + 1)^2 = 6^2
x^2 + (y + 1)^2 = 36