The equation of the circle that has a diameter with endpoints (12, 3) and(-18,3) is x² + y² + 6x - 6y - 207 = 0
Coordinates of center of circle
Since the endpoints of the diameter are (x₁, y₁) = (12,3) and (x₂, y₂) = (-18,3), the coordinates of the center of the circle are the midpoints of the diameter. So, the midpoints are
- x = (x₁ + x₂)/2 = (12 + (-18))/2 = (12 - 18)/2 = -6/2 = -3 and
- y = (y₁ + y₂)/2 = (3 + 3)/2 = 6/2 = 3.
So, the coordinates of the center of the circle are (-3, 3)
The radius of the circle
The radius of the circle r = √[(x₁ - h)² + (y₁ - k)²] where
- (x₁, y₁) = coordinates of end of diameter = (12, 3) and
- (h, k) = coordinates of center of circle = (-3, 3)
So, substituting the values of the variables into r, we have
r = √[(x₁ - h)² + (y₁ - k)²]
r = √[(12 - (-3))² + (3 - 3)²]
r = √[(12 + 3)² + 0²]
r = √[15² + 0²]
r = √15²
r = 15
The equation of the circle
The equation of a circle with center (h,k) is given by
(x - h)² + (y - k)² = r² where r = radius of circle.
Substituting the values of the variables into the equation, we have
(x - h)² + (y - k)² = r²
(x - (-3))² + (y - 3)² = 15²
(x + 3)² + (y - 3)² = 15²
x² + 6x + 9 + y² - 6y + 9 = 225
x² + 6x + y² - 6y + 9 + 9 - 225 = 0
x² + y² + 6x - 6y - 207 = 0
The equation of the circle that has a diameter with endpoints (12, 3) and(-18,3) is x² + y² + 6x - 6y - 207 = 0
Learn more about equation of a circle here:
https://brainly.com/question/18435467
