how do you determine the equation of a circle that has diameter endpoints (-2,7) and (2,-7)

the equation format is r^2 = x^2 + y^2

thanks

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how do you determine the equation of a circle that has diameter endpoints (-2,7) and (2,-7)

the equation format is r^2 = x^2 + y^2

thanks

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## Answers (3)

mid point of diameter, which is the centre of circle is (0, 0)

radius = √(2² + 7²) = √53

equation is

(√53)² = (x−0)² + (y−0)²

answer

53 = x² + y².

First, find the midpoint between the 2 points:

x = (-2 + 2) / 2 = 0/2 = 0

y = (7 - 7) / 2 = 0/2 = 0

So, the circle is centered at (0 , 0)

The radius is the distance from the center to any point on the surface of the circle

(-2 - 0)^2 + (7 - 0)^2 = r^2

4 + 49 = r^2

53 = r^2

(x - 0)^2 + (y - 0)^2 = 53

x^2 + y^2 = 53

it is....you can plot points, or just memorize the formula. -_-

Reference:Ting ting Li gave me credit for embarrassing her wall.