Find the equation of circle with the center and passes through the given point.
Problem 1 :
Center : (11, 0)
Point on Circle : (3, 0)
Solution :
Equation of a circle with center (h, k) and radius r :
(x - h)^{2} + (y - k)^{2} = r^{2}
Centre (h, k) = (11, 0)
(x - 11)^{2} + (y - 0)^{2} = r^{2} --- (1)
The given circle is passing through the point (3, 0).
Then substitute 3 for x and 0 for y.
(3 - 11)^{2} + (0 - 0)^{2} = r^{2}
(-8)^{2} = r^{2}
64 = r^{2}
Standard equation of a circle :
(x - 11)^{2} + (y - 0)^{2} = 64
x^{2} + (11)^{2} - 2(x) (11) + y^{2} = 64
x^{2} + 121 - 22x + y^{2} = 64
x^{2} + y^{2} - 22x + 121 = 64
Subtract 64 from each side.
General form of equation of a circle :
x^{2} + y^{2} - 22x + 57 = 0
Problem 2 :
Center : (15, 13)
Point on Circle : (19, 13)
Solution :
Equation of a circle with centre (h, k) and radius r :
(x - h)^{2} + (y - k)^{2} = r^{2}
Centre (h, k) = (15, 13)
(x - 15)^{2} + (y - 13)^{2} = r^{2} --- (1)
The given circle is passing through the point (19, 13).
Then substitute 19 for x and 13 for y.
(19 - 15)^{2} + (13 - 13)^{2} = r^{2}
(4)^{2} = r^{2}
16 = r^{2}
Standard equation of a circle :
(x - 15)^{2} + (y - 13)^{2} = 16
x^{2} + (15)^{2} - 2(x) (15) + y^{2} + (13)^{2} - 2(y)(13) = 16
x^{2} + 225 - 30x + y^{2} + 169 - 26y = 16
x^{2} + y^{2} - 30x - 26y + 394 = 16
Subtract 16 from each side.
General form of equation of a circle :
x^{2} + y^{2} - 30x - 26y + 378 = 0
Problem 3 :
Center : (-5, 9)
Point on Circle : (-7, 11)
Solution :
Equation of a circle with centre (h, k) and radius r :
(x - h)^{2} + (y - k)^{2} = r^{2}
Centre (h, k) = (-5, 9)
(x + 5)^{2} + (y - 9)^{2} = r^{2} --- (1)
The given circle is passing through the point (-7, 11).
Then substitute -7 for x and 11 for y.
(-7 + 5)^{2} + (11 - 9)^{2} = r^{2}
(-2)^{2} + (8)^{2} = r^{2}
4 + 64 = r^{2}
68 = r^{2}
Standard equation of a circle :
(x + 5)^{2} + (y - 9)^{2} = 68
x^{2} + (5)^{2} + 2(x) (5) + y^{2} + 9^{2} - 2(y)(9) = 68
x^{2} + 25 + 10x + y^{2} + 81 - 18y = 68
x^{2} + y^{2} + 10x - 18y + 106 = 68
Subtract 68 from each side.
General form of equation of a circle :
x^{2} + y^{2} + 10x - 18y + 38 = 0
Problem 4 :
Center : (-11, 11)
Point on Circle : (-15, 17)
Solution :
Equation of a circle with centre (h, k) and radius r :
(x - h)^{2} + (y - k)^{2} = r^{2}
Centre (h, k) = (-11, 11)
(x + 11)^{2} + (y - 11)^{2} = r^{2} --- (1)
The given circle is passing through the point (-15, 17).
Then substitute -15 for x and 17 for y.
(-15 + 11)^{2} + (17 - 11)^{2} = r^{2}
(-4)^{2} + (6)^{2} = r^{2}
16 + 36 = r^{2}
52 = r^{2}
Standard equation of a circle :
(x + 11)^{2} + (y - 11)^{2} = 52
x^{2} + (11)^{2} + 2(x) (11) + y^{2} + (11)^{2} - 2(y)(11) = 52
x^{2} + 121 + 22x + y^{2} + 121 - 22y = 52
x^{2} + y^{2} + 22x - 22y + 242 = 52
Subtract 52 from each side.
General form of equation of a circle :
x^{2} + y^{2} + 22x - 22y + 190 = 0
Problem 5 :
Find the equation of a circle where the center is at (2, -4), and the point (6, 1) rests on the circle.
Solution :
Center : (2, -4)
Point on Circle : (6, 1)
Equation of a circle with center (h, k) and radius r :
(x - h)^{2} + (y - k)^{2} = r^{2}
Centre (h, k) = (2, -4)
(x - 2)^{2} + (y + 4)^{2} = r^{2} --- (1)
The given circle is passing through the point (6, 1).
Then substitute 6 for x and 1 for y.
(6 - 2)^{2} + (1 + 4)^{2} = r^{2}
(4)^{2} + (5)^{2} = r^{2}
16 + 25 = r^{2}
41 = r^{2}
Standard equation of a circle :
(x - 2)^{2} + (y + 4)^{2} = 41
x^{2} + (2)^{2} - 2(x) (2) + y^{2} + (4)^{2} + 2(y)(4) = 41
x^{2} + 4 - 4x + y^{2} + 16 + 8y = 41
x^{2} + y^{2} - 4x + 8y + 20 = 41
Subtract 41 from each side.
General form of equation of a circle :
x^{2} + y^{2} - 4x + 8y - 21 = 0
Problem 6 :
Find the equation of a circle where the center is at (-2, 3), and the point (1, 4) rests on the circle.
Solution :
Center : (-2, 3)
Point on Circle : (1, 4)
Equation of a circle with centre (h, k) and radius r :
(x - h)^{2} + (y - k)^{2} = r^{2}
Centre (h, k) = (-2, 3)
(x + 2)^{2} + (y - 3)^{2} = r^{2} --- (1)
The given circle is passing through the point (1, 4).
Then substitute 1 for x and 4 for y.
(1 + 2)^{2} + (4 - 3)^{2} = r^{2}
(3)^{2} + (1)^{2} = r^{2}
9 + 1 = r^{2}
10 = r^{2}
Standard equation of a circle :
(x + 2)^{2} + (y - 3)^{2} = 10
x^{2} + (2)^{2} + 2(x) (2) + y^{2} + (3)^{2} - 2(y)(3) = 10
x^{2} + 4 + 4x + y^{2} + 9 - 6y = 10
x^{2} + y^{2} + 4x - 6y + 13 = 10
Subtract 10 from each side.
General form of equation of a circle :
x^{2} + y^{2} + 4x - 6y + 3 = 0
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM