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What is the equation of a circle whose diameter end points are at 10 -4 and 2 2 on the Cartesian plane?
Wiki User
∙ 7y ago
End points: (10, -4) and (2, 2)
Midpoint: (6, -1)
Distance from (6, -1) to (10, -4) = 5
Distance from (6, -1) to (2, 2) = 5
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's end points are at 2 -3 and 8 7?
The mid point of the diameter is the circle's centre: Mid point of (2, -3) and (8, 7) is ((2 + 8)/2, (-3 + 7)/2) = (5, 2) The radius is half the diameter, but as the equation for a circle requires the radius squared: radius = diameter/2 → radius² = diameter² / 4 Diameter (and diameter²) can be found using Pythagoras: diameter² = change_in_x² + change_in_y² = (8 - 2)² + (7 - -3)² = 6² + 10² = 136 → radius² = 136/4 = 34 Circle with centre (X, Y) and radius r has equation: (x - X)² + (y - Y)² = r² → (x - 5)² + (y - 2)² = 34 → x² -10x + 25 + y² - 4y + 4 = 34 → x² + y² - 10x - 4x - 5 = 0
A straight line going through the center of a circle connecting two points circumference?
The diameter.
What are the uses of plotting points in cartesian plane?
They are the coordinates of x and y to plot out straight lines or curves on the Cartesian plane
What is used to locate and graph points in the Cartesian Coordinate System?
A pair of two points (2D) or 3 points (3D) written as (x,y) or (x,y,z).
What is the equation and its distance from origin that meets the line joining the points 13 17 and 19 23 at its midpoint on the Cartesian plane showing work?
Points: (13, 17) and (19, 23) Midpoint: (16, 20) Slope of required equation: 5/4 Its equation: 4y = 5x or as y = 1.25x Its distance from (0, 0) to (16, 20) = 4 times sq rt 41
What is the Cartesian equation of a circle whose end points of its diameter are at 10 -4 and 2 2?
End points: (10, -4) and (2, 2) Midpoint: (6, -1) which is the centre of the circle Distance from (6, -1) to any of its end points = 5 which is the radius Therefore the Cartesian equation is: (x-6)^2 +(y+1)^2 = 25
What is the equation of a circle whose diameter end points are at 8 7 and 2 -3 on the Cartesian plane?
End points: (8, 7) and (2, -3) Midpoint: (5, 2) which is the center of the circle Radius: square root of 34 Equation of the circle: (x-5)^2 +(y-2)^2 = 34
What is the equation and area of a circle whose diameter end points are at 2 -3 and 8 7 on the Cartesian plane showing work?
Diameter end points: (2, -3) and (8, 7) Centre of circle: (5, 2) Length of diameter: 2 times square root of 34 Equation: (x-5)^2+(y-2)^2 = 34 which in effect is the radius squared Area in square units: 34*pi
What is the equation of a circle whose diameter end points are at 2 -3 and 8 7 on the Cartesian plane?
Diameter endpoint's: (2, -3) and (8, 7) Midpoint: (5, 2) which is the centre of the circle Distance from (5, 2) to (2, -3) or (8, 7) = square root of 34 which is the radius Equation of the circle: (x-5)^2 +(y-2)^2 = 34
What of a circle is a chord that has the center of the circle as one of its points?
The diameter of the circle.
What is the radius of a circle and its centre that passes through the points of 5 0 and 3 4 and -5 0 on the Cartesian plane?
Points: (5, 0) and (3, 4) and (-5, 0) Equation works out as: x^2+y^2 = 25 Radius: 5 units in length Centre of circle is at the point of origin (0, 0) on the Cartesian plane.
what are endpoints on the diameter?
The points where a diameter intersects the circle are its endpoints.
What is the equation of the line whose end points are 5 2 and 3 2 on the Cartesian plane?
The equation is y = 2
What is a line segment that contains 2 points on a circle?
It is a chord and the largest chord in a circle is its diameter
What is a segment from two points on a circle which passes through the center of the circle?
It is the circle's diameter
What is the radius of a circle and its equation when its diameter touches the points of 2 -3 and 8 7 on the Cartesian plane showing key stages?
Points: (2, -3) and (8, 7) Centre: (8+2)/2 and (7-3)/2 = (5, 2) Radius: (8-5)2+(7-2)2 = 34 and the square root of this is the radius Equation: (x-5)2+(y-2)2 = 34
What is the centre and radius of a circle enscribed through the points of 6 2 and 8 -2 and 0 2 on the Cartesian plane?
It works out that the circle's centre is at (3, -2) and its radius is 5 on the Cartesian plane.
