A circle cannot have end points!
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
centre is the midpoint of the line → centre = ((-2 + 6)/2, (8 + 4)/2) = (2, 6)
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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The equation works out as: (x-1)2+(y-4)2 = 25 whereas (1, 4) is the circle's centre and 25 is the radius2
Diameter endpoints: (2, 7) and (-6, -1) Centre of circle will be the midpoint which is: (-2, 3)
Its center is at (-1, 1/2)
Endpoints of diameter: (-10, -2) and (4, 6)Midpoint which is the center of the circle: (-3, 2)Radius of the circle: square root of 65Equation of the circle: (x+3)^2 +(y-2)^2 = 65
Diameter is the square root of (1--7)2+(3--1)2 = 4 times sq rt of 5 Circles center coordinates: (-3, 1)
Endpoints: (10, -4) and (2, 2) Midpoint: (6, -1) which is the centre of the circle Distance from (6, -1) to (10, -4) or (2, 2): 5 which is the radius of the circle Therefore equation of the circle: (x-6)^2 + (y+1)^2 = 25
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
Yes, it have.
First, find the coordinates of the diameter midpoint. For example, the endpoints of a diameter have coordinates of (x1, y1) = (1, 2) and (x2, y2) = (7, 6). By the midpoint formula we have: (xm, ym) = [(x2 - x1)/2, (y2 - y1)/2] = [(7 - 1)/2, (6 - 2)/2] = (3, 2) Since (x1, y1) it is the closest point to the origin, the coordinates of the center of the circle are: (xc, yc) = (xm + x1, ym + y1) = (3 + 1, 2 + 2) = (4, 4).
1. Find the coordinates of the center of the circle. Call it point (a, b). To find this point, calculate the average of the x-coordinates of the endpoints, and also the average of the y-coordinates. 2. Find the radius of the circle. Use the formula for distance (which is based on Pythagoras' Theorem). Call the length of the radius "r". 3. The formula for the circle is (x - a)2 + (y - b)2 = r2. Replace the values you found earlier.
centre is the midpoint of the line → centre = ((-2 + 6)/2, (8 + 4)/2) = (2, 6)
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
dinosaurs