To find the equation of a circle with center (-3, 0) containing the point (-1, √104), we first need to find the radius of the circle.
The distance between the center (-3, 0) and the point (-1, √104) is the radius of the circle.
Using the distance formula:
r = √((x2 - x1)^2 + (y2 - y1)^2)
r = √((-1 - (-3))^2 + (√104 - 0)^2)
r = √(2^2 + 104)
r = √(4 + 104)
r = √108
r = 6√3
Therefore, the radius of the circle is 6√3.
The equation of a circle with center (h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2
Substitute the center (-3, 0) and radius 6√3:
(x + 3)^2 + y^2 = (6√3)^2
(x + 3)^2 + y^2 = 108
The equation of the circle is:
(x + 3)^2 + y^2 = 108
Determine the equation of the circle with center left bracket, minus, 3, comma, 0, right bracket(−3,0) containing the point left bracket, minus, 1, comma, square root of, 104, end square root, right bracket(−1,
104
).
1 answer