0
What is the radius the centre and its equation of a circle passing though the points 5 5 and 8 5 and 5 1 on the Cartesian plane showing work?
Wiki User
∙ 7y ago
Use formula: x^2 +2gx +y^2 +2fy +c = 0
Points: (5, 5) (8, 5) (5, 1)
From the above information form three simultaneous equations as follows:-
10g +10f +c = -50
16g +10f +c = -89
10g +2f +c = -26
Solving the simultaneous equations: g = -6.5, f = -3 and c = 45
So: x^2 -13x +y^2 -6y +45 = 0
Completing the squares: (x-6.5)^2 +(y-3)^2 -42.25 -9 +45 = 0
Equation of the circle: (x-6.5)^2 +(y-3)^2 = 6.25
Centre of circle: (6.5, 3)
Radius of the circle: square root of 6.25 = 2.5 units
Wiki User
∙ 7y ago
Wiki User
∙ 7y agoRadius: 2.5Centre: (6.5, 3)
Equation: (x - 6.5)^2 + (y - 3)^2 = 6.25
Add your answer:
Earn +20 pts
What is the centre and radius of a circle passing through the points of 5 5 and 8 5 and 5 1 on the Cartesian plane?
Using the formula of x^2 +2gx +y^2 +2fy +c = 0 it works out that the centre of the circle is at (6.5, 3) and its radius is 2.5 units in length. Alternatively plot the points on the Cartesian plane to find the centre and radius of the circle.
What is the centre and radius of a circle embeded on the Cartesian plane whose equation is x squared plus y squared -4x -2y -4 equals 0 showing key aspects of work?
Note that: (x-a)2+(y-b)2 = radius2 whereas a and b are the coordinates of the circle's centre Equation: x2+y2-4x-2y-4 = 0 Completing the squares: (x-2)2+(y-1)2 = 9 Therefore: centre = (2, 1) and radius = 3
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 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 centre and radius of a circle passing through the points of 6 3 and -5 2 and 7 2 on the Cartesian plane?
Points: (6, 3) (-5, 2) and (7, 2) Circle's equation works out as: (x-1)^2+(y+3)^2 = 61 Centre of the circle is at: (1, -3) Radius of the circle is the square root of 61 which is about 7.81 to two decimal places
What is the Cartesian equation of a circle whose centre is in the 1st quadrant with a radius of 7 and touches the x and y axes?
Centre of the circle is at (7, 7) and its Cartesian equation is (x-7)^2 + (y-7)^2 = 49
What is the centre and radius of a circle passing through the points of 5 5 and 8 5 and 5 1 on the Cartesian plane?
Using the formula of x^2 +2gx +y^2 +2fy +c = 0 it works out that the centre of the circle is at (6.5, 3) and its radius is 2.5 units in length. Alternatively plot the points on the Cartesian plane to find the centre and radius of the circle.
What is the equation of a circle with center (3 8) and a radius of 2?
Centre of the circle: (3, 8) Radius of the circle: 2 Cartesian equation of the circle: (x-3)^2 + (y-8)^2 = 4
What is the centre and radius of a circle embeded on the Cartesian plane whose equation is x squared plus y squared -4x -2y -4 equals 0 showing key aspects of work?
Note that: (x-a)2+(y-b)2 = radius2 whereas a and b are the coordinates of the circle's centre Equation: x2+y2-4x-2y-4 = 0 Completing the squares: (x-2)2+(y-1)2 = 9 Therefore: centre = (2, 1) and radius = 3
What is the equation of a circle whose centre is at 3 -5 and meets the point of 6 -7 on the Cartesian plane?
Centre of circle: (3, -5) Distance from (3, -5) to (6, -7) is the square root of 13 which is the radius Equation of the circle: (x-3)^2 + (y+5)^2 = 13
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?
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
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 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 centre and radius of a circle passing through the points of 6 3 and -5 2 and 7 2 on the Cartesian plane?
Points: (6, 3) (-5, 2) and (7, 2) Circle's equation works out as: (x-1)^2+(y+3)^2 = 61 Centre of the circle is at: (1, -3) Radius of the circle is the square root of 61 which is about 7.81 to two decimal places
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.
What is the equation of the straight line that extends from the origin to the centre of the points 3 2 and 5 -1 on the Cartesian plane giving thorougher details of your work?
The equation is y = 1/8x because there is no y intercept and by doing some homework you'll find it correct
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
