Question

Write an equation of the circle that has a diameter with endpoints (12, 3) and (-18,3).

Answer 1

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

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