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What is the equation of a circle whose endpoints of its diameter are at 2 2 and 10 -4 on the Cartesian plane showing work?
Wiki User
∙ 7y ago
Endpoints: (2, 2) and (10, -4)
Midpoint: (6, -1) which is the centre of the circle
Distance from (6, -1) to (2, 2) or (10, -4) = 5 which is the radius of the circle
Therefore equation of the circle: (x-6)^2 + (y+1)^2 = 25
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∙ 7y ago
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What is the equation of a circle whose diameter endpoints are at 10 -4 and 2 2 on the Cartesian plane and what are the perpendicular equations to its endpoints showing work?
Endpoints of diameter: (10, -4) and (2, 2)Midpoint which is the center of the circle: (6, -1)Distance from (10, -4) or (2, 2) to (6, -1) = 5 which is the radius of the circleEquation of the circle: (x-6)^2 +(y+1)^2 = 25Slope of radius: -3/4Slope of perpendicular equations which will be parallel: 4/31st perpendicular equation: y--4 = 4/3(x-10) => 3y = 4x-522nd perpendicular equation: y-2 = 4/3(x-2) => 3y = 4x-2
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What is the equation of a circle whose diameter endpoints are at 10 -4 and 2 2 on the Cartesian plane and what are the perpendicular equations to its endpoints showing work?
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