To find equation of the circle with endpoints of diameter, we use the formula given below.
Equation of circle :
(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0
Problem 1 :
Ends of a diameter : (-9, 5) and (1, 1)
Solution :
Now we have to consider the given points as (x_{1}, y_{1}) and (x_{2}, y_{2}). So the values of x_{1} = -9, y_{1} = 5, x_{2} = 1 and y_{2} = 1
Equation of circle :
(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0
(x + 9) (x - 1) + (y - 5) (y - 1) = 0
x^{2} - x + 9x - 9 + y^{2} - y - 5y + 5 = 0
x^{2} + 8x - 4 + y^{2} - 6y = 0
So, the required equation of a circle
x^{2} + y^{2} + 8x - 6y - 4 = 0
Problem 2 :
Ends of a diameter : (6, 8) and (16, 6)
Solution :
Here x_{1} = 6, y_{1} = 8, x_{2} = 16 and y_{2} = 6
Equation of circle :
(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0
(x - 6) (x - 16) + (y - 8) (y - 6) = 0
x^{2} - 16x - 6x + 96 + y^{2} - 6y - 8y + 48 = 0
x^{2} - 22x + 144 + y^{2} - 14y = 0
So, the required equation of a circle
x^{2} + y^{2} - 22x - 14y + 144 = 0
Problem 3 :
Ends of a diameter : (19, 5) and (-5, 1)
Solution :
Here x_{1} = 19, y_{1} = 5, x_{2} = -5 and y_{2} = 1
(x - 19) (x + 5) + (y - 5) (y - 1) = 0
x^{2} + 5x - 19x - 95 + y^{2} - y - 5y + 5 = 0
x^{2} - 14x - 90 + y^{2} - 6y = 0
So, the required equation of a circle
x^{2} + y^{2} - 14x - 6y - 90 = 0
Problem 4 :
Ends of a diameter : (11, -5) and (11, -7)
Solution :
Here x_{1} = 11, y_{1} = -5, x_{2} = 11 and y_{2} = -7
(x - 11) (x - 11) + (y + 5) (y + 7) = 0
x^{2} - 11x - 11x + 121 + y^{2} + 7y + 5y + 35 = 0
x^{2} - 22x + 156 + y^{2} + 12y = 0
So, the required equation of a circle
x^{2} - 22x + 156 + y^{2} + 12y = 0.
Problem 5 :
Given a circle with (5, 1) and (3, -1) as the endpoints of the diameter.
Solution :
Here x_{1} = 5, y_{1} = 1, x_{2} = 3 and y_{2} = -1
(x - 5) (x - 3) + (y - 1) (y + 1) = 0
x^{2} - 3x - 5x + 15 + y^{2} + y - y - 1 = 0
x^{2} - 8x + 14 + y^{2} = 0
So, the required equation of a circle x^{2} + y^{2} - 8x + 14 = 0.
Problem 6 :
Given a circle with (2, 1) and (6, -3) as the endpoints of the diameter.
Solution :
Here x_{1} = 2, y_{1} = 1, x_{2} = 6 and y_{2} = -3
(x - 2) (x - 6) + (y - 1) (y + 3) = 0
x^{2} - 6x - 2x + 12 + y^{2} + 3y - y - 3 = 0
x^{2} - 8x + 9 + y^{2} + 2y = 0
So, the required equation of a circle
x^{2} + y^{2} - 8x + 2y + 9 = 0.
Problem 7 :
Given a circle with (4, -3) and (2, 1) as the endpoints of the diameter.
Solution :
Here x_{1} = 4, y_{1} = -3, x_{2} = 2 and y_{2} = 1
(x - 4) (x - 2) + (y + 3) (y - 1) = 0
x^{2} - 2x - 4x + 8 + y^{2} - y + 3y - 3 = 0
x^{2} - 6x + 5 + y^{2} + 2y = 0
So, the required equation of a circle
x^{2} + y^{2} - 6x + 2y + 5 = 0.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM